The remaining species are hydroxide ions (H+) and sulfide ions (S2-).
Thus, the net ionic equation for the reaction of excess hydrobromic acid and calcium sulfide is as follows:
2H+ (aq) + S2- (aq) → H2S (g)
1. Net ionic equation for the reaction of excess hydrobromic acid and potassium carbonate:
Hydrobromic acid reacts with potassium carbonate as per the following balanced equation:
HBr + K2CO3 → 2KBr + H2O + CO2
The complete ionic equation is as follows:
2H+ (aq) + 2Br- (aq) + 2K+ (aq) + CO32- (aq) → 2K+ (aq) + 2Br- (aq) + H2O (l) + CO2 (g)
Here, the spectator ions are K+ and Br-, which are canceled out from both the sides of the equation. The remaining species are hydroxide ions (H+) and carbonate ions (CO32-).
Thus, the net ionic equation for the reaction of excess hydrobromic acid and potassium carbonate is as follows:
2H+ (aq) + CO32- (aq) → H2O (l) + CO2 (g)
2. Net ionic equation for the reaction of excess hydrobromic acid and calcium sulfide:
Hydrobromic acid reacts with calcium sulfide as per the following balanced equation:
HBr + CaS → CaBr2 + H2S
The complete ionic equation is as follows:
2H+ (aq) + 2Br- (aq) + CaS (s) → CaBr2 (aq) + H2S (g)
Here, the spectator ions are Ca2+ and Br-, which are canceled out from both the sides of the equation. The remaining species are hydroxide ions (H+) and sulfide ions (S2-).
Thus, the net ionic equation for the reaction of excess hydrobromic acid and calcium sulfide is as follows:
2H+ (aq) + S2- (aq) → H2S (g)
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Draw a Lewis structure for the molecule below, showing all lone pairs. You may abbreviate any methyl g CH 2
CHCHBrCH 3
The Lewis dot structure is given below in the image.
Lewis dot structure, often referred to as electron dot structure or Lewis structure, is a diagram that represents a molecule or an ion and displays how its atoms and valence electrons are arranged.
Each atom is represented by its chemical symbol in a Lewis dot structure, and the valence electrons are shown as dots or dashes. The sign is surrounded by dots, each of which stands for a valence electron.
In order to establish a stable electron configuration with eight valence electrons, it is necessary to distribute the valence electrons in a fashion that satisfies the octet rule, which stipulates that atoms typically gain, lose, or share electrons. The total number of valence electrons for all the atoms must be known in order to draw a Lewis dot structure.
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Malonates must bond at what specific location on an enzyme, and are considered what kind of molecule?
Malonates must bond at active site on an enzyme, and are considered enzyme inhibitor.
Malonate is a dicarboxylic acid with three carbons. It is well known as a succinate dehydrogenase competitive inhibitor.
It naturally occurs in biological systems, such as developing rat brains and legumes, indicating that it may be crucial for symbiotic nitrogen metabolism and brain growth.
Malonate ions, which closely resemble succinate ions in structure, prevent succinate dehydrogenase from completing the conversion. The malonate ions' similar shapes enable them to connect to the active site, but the absence of the CH2-CH2 link in the middle of the ion prevents any further reaction from occurring.
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Lquid hexane (CH 3
(CH 2
) 4
CH 3
) reacts with gaseous axygen gas (O 2
) to produce gaseous carbon dioxide (CO 2
) and gaseous water (H 2
O), If 34.6 g of water is produced from the reaction of 52.56 g of hexane and 315.2 g of axygen gas, calculate the percent yeld of water. Round your answer to 3 significant figures.
The percent yield of water is 98.07%.Given: 34.6 g of H2O, 52.56 g of hexane (C6H14), and 315.2 g of oxygen gas (O2)Reacting hexane with oxygen gas gives CO2 and H2O.
The balanced chemical reaction for the given reaction is as follows:
2C6H14(l) + 19O2(g) → 14CO2(g) + 14H2O(l)
Molar mass of hexane (C6H14) = 6(12.01 g/mol) + 14(1.01 g/mol) = 86.18 g/mol.
Mass of 52.56 g of hexane is given, we will find the number of moles of hexane. n(hexane) = (52.56 g) / (86.18 g/mol) = 0.609 mol
Molar mass of oxygen (O2) = 2(16.00 g/mol) = 32.00 g/mol. Number of moles of oxygen can be calculated as follows:
n(O2) = (315.2 g) / (32.00 g/mol) = 9.85 mol
Mole ratio of H2O to hexane is 7:2.
Therefore, the number of moles of water can be calculated as follows:
n(H2O) = (7/2) × n(hexane) = (7/2) × 0.609 = 2.126 mol
Mass of water that should be produced based on the number of moles of hexane consumed is given by multiplying the number of moles of hexane by the molar mass of water (18.02 g/mol).
Mass of water that should be produced = (0.609 mol) × (7/2) × (18.02 g/mol) = 35.28 gPercent yield of water can be calculated as follows:
Percent yield = (Actual yield / Theoretical yield) × 100We are given the actual yield of water.
Therefore, the percent yield can be calculated as follows:
Percent yield = (34.6 g / 35.28 g) × 100 = 98.07%
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Be sure to answer all parts. How many H atoms are in 42.7 g of isopropanol (rubbing alcohol), C 3
H 8
O ? Enter your answer in scientific notation. ×10 H atoms
The mass of isopropanol is given as 42.7 g. The molar mass of isopropanol can be calculated as:
Molar mass (C3H8O) = 12.01 × 3 + 1.01 × 8 + 16.00 = 60.09 g/mol
The number of moles of isopropanol can be calculated as:
Number of moles = mass/molar mass = 42.7/60.09 = 0.7111 mol
Using the coefficients in the balanced chemical equation for the combustion of isopropanol:
C3H8O + 5 O2 → 4 H2O + 3 CO2
We can see that there are 4 H atoms in every molecule of isopropanol.
The total number of H atoms in 0.7111 mol of isopropanol can be calculated as:
Number of H atoms = 4 × Avogadro's number × number of moles
= 4 × 6.022 × 1023 × 0.7111= 1.712 × 1024
Therefore, there are 1.712 × 1024 H atoms in 42.7 g of isopropanol (C3H8O) in scientific notation.
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The boiling point of diethyl ether, CH 3
CH 2
OCH 2
CH 3
, is 34.500 ∘
C at 1 atmosphere. K b
( diethyl ether )=2.02 ∘
C/m In a laboratory experiment, students synthesized a new compound and found that when 11.69 grams of the compound were dissolved in 251.8 grams of diethyl ether, the solution began to boil at 34.765 ∘
C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound? g/mol 9 more group attempts remaining The freezing point of benzene, C 6
H 6
, is 5.500 ∘
C at 1 atmosphere. K f
(benzene) =5.12 ∘
C/m In a laboratory experiment, students synthesized a new compound and found that when 14.67 grams of the compound were dissolved in 271.1 grams of benzene, the solution began to freeze at 4.718 ∘
C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound? g/mol 9 more group attempts remaining
The molecular weight of the compound is 88 g/mol.
Molecular weight of the compound given the boiling point of diethyl ether, CH3CH2OCH2CH3, and the freezing point of benzene, C6H6 is 88 g/mol.
For this particular problem, the given values of the boiling point of diethyl ether and freezing point of benzene are required to be utilized.
1. For boiling point elevation, ΔTb = Kb x molality.
The molality can be calculated as:
(11.69g)/(134.7 g/mol) = 0.0867 mol Diethyl ether [mass / molar mass]
Thus, ΔTb = (2.02°C/m) × (0.0867 mol/kg)
= 0.175°C.
The boiling point of the solution = (34.765 + 0.175)°C
= 34.94°C.
For this equation, the unknown value is the molecular weight of the compound and therefore it can be calculated as:
Molecular weight = (1000 g/kg) × [(1.015 × 34.5°C)/(2.02°C/m × 0.2518 kg) - 1] × (134.7 g/mol)
The answer of which comes out to be, 88 g/mol.
2. For freezing point depression, ΔTf = Kf x molality.
The molality can be calculated as:
(14.67 g) / (molar mass of compound) + (271.1 g)
= 0.10 mol/kg Benzene [mass / molar mass]
ΔTf = (5.12°C/m) × (0.10 mol/kg)
= 0.512°C.
The freezing point of the solution = (5.50 - 0.512)°C
= 4.988°C.
Now, the unknown value in this equation is the molecular weight of the compound and therefore it can be calculated as:
Molecular weight = (1000 g/kg) × [(5.50°C - 4.718°C)/(5.12°C/m × 0.2711 kg) ] × (78.1 g/mol)
The answer comes out to be 88 g/mol. Therefore, the molecular weight of the compound is 88 g/mol.
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according to molecular orbital theory, two separate px orbitals interact about the x-axis to form what molecular orbitals?
When two separate pₓ orbitals interact about the x-axis, they form a bonding molecular orbital (σ bonding) and an antibonding molecular orbital (σ* antibonding). The bonding orbital promotes electron sharing and contributes to the stability of the molecular bond, while the antibonding orbital weakens the bond.
According to molecular orbital theory, two separate pₓ orbitals can interact about the x-axis to form two molecular orbitals: a bonding molecular orbital (σ bonding) and an antibonding molecular orbital (σ* antibonding).
When two pₓ orbitals interact, they combine in-phase to form a bonding molecular orbital. In this case, the wave functions of the two pₓ orbitals align constructively, resulting in a region of electron density between the two nuclei. This bonding molecular orbital has lower energy than the original pₓ orbitals and promotes electron sharing between the two atoms.
On the other hand, the out-of-phase combination of the pₓ orbitals results in an antibonding molecular orbital. The wave functions of the pₓ orbitals align destructively, leading to a region of electron density with opposite signs on each atom. This antibonding molecular orbital has higher energy than the original pₓ orbitals and does not promote electron sharing between the atoms.
The bonding molecular orbital is denoted as σ bonding, while the antibonding molecular orbital is denoted as σ* antibonding. The σ bonding orbital is more stable and lower in energy, contributing to the formation of a stable molecular bond. The σ* antibonding orbital is less stable and higher in energy, and it weakens the overall bonding between the atoms.
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The mass spectrum of 3-Methyl-4-phenyl-2-butanone showed the following positive fragment ions at M/Z: 162, 147, 43, and 91. Draw the structures of the fragment positive ions that correspond to each mass above. [Note: each must bear a positive charge; you won't receive credit for neutral or negative fragments, even if it corresponds to the mass]
Thus, the mass spectrum of 3-Methyl-4-phenyl-2-butanone showed the following positive fragment ions at M/Z: 162, 147, 43, and 91.
The mass spectrum of 3-Methyl-4-phenyl-2-butanone showed the following positive fragment ions at M/Z: 162, 147, 43, and 91.
The positive fragment ions corresponding to each mass are as follows:
At m/z = 162, the ion corresponds to the loss of a CO (28) unit from the parent molecule.
The positive fragment ion structure for m/z = 162 is shown below:
At m/z = 147, the ion corresponds to the loss of a CH3CH2CH2 (44) unit from the parent molecule.
The positive fragment ion structure for m/z = 147 is shown below:
At m/z = 43, the ion corresponds to the loss of a C7H7COCH3 (119) unit from the parent molecule.
The positive fragment ion structure for m/z = 43 is shown below:
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In the coal-gasification process, carbon monoxide is converted to carbon dioxide vi the following reaction: CO(g)+H 2
O(g)⇌CO 2
( g)+H 2
( g) In an experiment, 0.35 mol of CO and 0.40 mol of H 2
O were placed in a 1.00−L reaction vessel. At equilibrium, there were 0.19 mol of CO remaining. K eq
at the temperature of the experiment is 1.78 0.56 1.0 0.75 5.47 The equilibrium constant for reaction 1 is K. The equilibrium constant for reaction 2 is (1) SO 2
( g)+(1/2)O 2
( g)⇌SO 3
( g) (2) 2SO 3
( g)⇌2SO 2
( g)+O 2
( g) 1/2 K 2 K 1/K 2
−K 2
K 2
Partial pressure of oxygen qubed over partial pressure of ozone squared 2HI(g)⇌H 2
( g)+I 2
( g) H 2
( g)+Cl 2
( g)⇌2HCl(g) N 2
( g)+3H 2
( g)⇌2NH 3
( g) 2SO 3
( g)⇌2SO 2
( g)+O 2
( g) 2Fe 2
O 3
( s)⇌4Fe(s)+3O 2
( g)
It is related to the equilibrium constant Kc 1 for the reaction SO2(g) + 1/2O2(g) ⇌ SO3(g) by the following equation:
Kc = (Kc1)3 / (4Kc2) = (1/4) (Kc1)3 / (Kc2)where Kc2 = [SO2]2 [O2] / [SO3]2Kc1 = [SO3] / ([SO2] [O2]1/2)
Therefore, the answer is 1/Kc1.
In the given chemical reaction of coal-gasification process, CO gas is converted to CO2 by the given equation:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
This reaction at equilibrium is represented as follows:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Initial molar concentration (I)0.35 mol/L of CO(g) and 0.40 mol/L of H2O(g) in 1.00 L of reaction vessel Equilibrium concentration (E)
The molar concentration of CO(g) at equilibrium
= 0.19 mol/L Let x be the change in molar concentration.
The molar concentration of the other components are as follows:
Concentration of CO(g)
= 0.35 - x Concentration of H2O(g)
= 0.40 - x Concentration of CO2(g)
= x Concentration of H2(g)
= x
The equilibrium constant Kc for the reaction
CO(g) + H2O(g) ⇌ CO2(g) + H2(g) can be expressed as follows:
Kc = [CO2] [H2] / [CO] [H2O]
= x * x / (0.35 - x) (0.40 - x)
Substitute the values in the expression and simplify it:1.78
= x2 / (0.35 - x) (0.40 - x)x2
= 1.78 (0.35 - x) (0.40 - x)x2
= 1.78 (0.14 - 0.75x + x2)x2 - 1.78 x2 + 1.33 x - 0.0504
= 0
Solve the quadratic equation, we get the value of x as follows:x = 0.157 M
Therefore, the concentration of CO2 at equilibrium is 0.157 M.
The equilibrium constant of the reaction
2SO3(g) ⇌ 2SO2(g) + O2(g) is Kc
= [SO2]2 [O2] / [SO3]2.
It is related to the equilibrium constant Kc 1 for the reaction
SO2(g) + 1/2O2(g) ⇌ SO3(g) by the following equation:Kc
= (Kc1)3 / (4Kc2)
= (1/4) (Kc1)3 / (Kc2)where Kc2
= [SO2]2 [O2] / [SO3]2Kc1
= [SO3] / ([SO2] [O2]1/2)
Therefore, the answer is 1/Kc1.
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for the spectrophotometer experiment, you will be making five dilutions of [ select ] from a stock solution with a concentration of [ select ] . what is the formula you use to make the dilutions?
To make dilutions for the spectrophotometer experiment, you will need to use the formula: C1V1 = C2V2 C1 represents the initial concentration of the stock solution, V1 represents the initial volume of the stock solution, C2 represents the desired concentration of the diluted solution, V2 represents the final volume of the diluted solution.
To make the dilutions, you will need to determine the desired concentration for each diluted solution. Let's say you want to make five dilutions. You will start with a stock solution, which has a known concentration. For each dilution, you will use the formula C1V1 = C2V2 to calculate the volumes required. First, you will select the volume of the stock solution you want to use (V1).
Then, you will select the desired concentration for the diluted solution (C2). Next, you will calculate the volume of the diluted solution needed (V2). For example, if you want to dilute the stock solution by a factor of 10, you would divide the initial concentration (C1) by 10 to get the desired concentration (C2). Finally, using the formula C1V1 = C2V2, you would solve for V2. Once you have the volume of the diluted solution, you can add the appropriate amount of solvent to reach the desired volume. Repeat this process for each dilution, adjusting the desired concentration and volumes accordingly.
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A student sets up the following equation to convert a measurement. (The? stands for a number the student is going to calculate.) Fill in the missing part of this equation. (0.070 mL)⋅=?dL
The answer is that 0.070 milliliters is equivalent to 0.070 deciliters. The missing part of the equation is to determine what value should be multiplied by 0.070 mL to convert it to deciliters (dL).
In this case, 1 deciliter (dL) is equivalent to 100 milliliters (mL). Therefore, to convert mL to dL, the student needs to multiply the given measurement of 0.070 mL by the conversion factor of 1 dL/100 mL.
To calculate the result, the student would set up the equation as follows:
(0.070 mL) * (1 dL/100 mL) = ? dL
Now, let's explain the answer. The conversion factor of 1 dL/100 mL is derived from the relationship between milliliters and deciliters. Since 1 deciliter is equal to 100 milliliters, we express this relationship as 1 dL/100 mL. When multiplying 0.070 mL by this conversion factor, the milliliters cancel out, leaving us with the result in deciliters. The calculation would be:
0.070 mL * (1 dL/100 mL) = 0.070 dL
Therefore, the answer is that 0.070 milliliters is equivalent to 0.070 deciliters.
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The+mineral+hausmannite+is+a+compound+of+55mn+and+16o.+if+72%+of+the+mass+of+hausmannite+is+due+to+manganese,+what+is+the+empirical+formula+of+hausmannite?
The empirical formula of the given mineral hausmannite that is a compound of manganese-55 and oxygen-16 is: Mn₃O₄
How to calculate the Empirical Formula?The parameters in the given question are:
Percentage of Manganese (Mn) is: 72%
Percentage of Oxygen (O) is: 100 – 72 = 28%
Molar mass of Mn is 55 and Molar Mass of Oxygen is 16. Thus:
Ratio of Mn = 72 / 55 = 1.309
Ratio of O = 28 / 16 = 1.75
Divide by the smallest to get:
Mn = 1.309 / 1.309 = 1
O = 1.75 / 1.309 = 1.34
Multiply by 3 to express in whole number
Mn = 1 × 3 = 3
O = 1.34 × 3 ≈ 4
Thus, the empirical formula is: Mn₃O₄
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Complete Question is:
The mineral hausmannite is a compound of manganese-55 and oxygen-16. If 72% of the mass of hausmannite is due to manganese, what is the empirical formula of Hausmannite? socratic.org
[11] Write the nuclear symbol for the atom with the following subatomic particles: 53p +
,54n,53e −
.
There are 53 protons because the atomic number is equal to the number of protons.
The nuclear symbol for the atom with the following subatomic particles: 53p+, 54n, 53e- can be written as follows:
53 is the atomic number because it has 53 protons, and protons' charge is +1.
The mass number of the atom is 107 because it has 53 protons (53 x 1) + 54 neutrons (54 x 1) = 107.
The symbol of the element is X, where X is the symbol of the element as found on the periodic table.
Hence the symbol for this atom is:
X10753
There are 53 protons because the atomic number is equal to the number of protons.
The number of neutrons can be calculated by subtracting the atomic number from the mass number.
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(4pts) Determination of the Mass Percent of NH4Cl Recovered from the Mixture Use your data to make the necessary calculations. Be sure to report answers with the correct number of significant figures. Mass of evaporating dish #1: Mass of evaporating dish #1 and original sample: 38.120 g 39.070 g (1pts) Mass of original sample ( g) Mass of evaporating dish #1 and sample after subliming NH 4
: 38.944 g (1pts) Mass of NH 4
Cl(g) (2pts) Percent by mass of NH 4
Cl(%) (3pts) Determination of the Mass of NaCl Recovered from the Mixture Mass of evaporating dish #2: Mass of watch glass: Mass of evaporating dish #2, watch glass, and NaCl : (1pts) Mass of NaCl(g) (2pts) Percent by mass of NaCl(%) (5pts) Select a reasonable explanation to account for the differences based on your data and results. There may be more than one possible reason that makes sense, but just select one of them. A. It is possible not all of the water was evaporated from the sand, causing the experimental mass to be higher. B. It is possible not all of the water was evaporated from the sand, causing the experimental mass to be lower. C. While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the experimental mass to be lower. D. While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the experimental mass to be higher. E. There was no difference in recovered and original mass, so there is no difference to account for.
While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the experimental mass to be lower. Hence option C is the correct answer.
Given data, Mass of evaporating dish
#1: 38.120 g Mass of evaporating dish #1 and original sample: 39.070 g Mass of evaporating dish #1 and sample after subliming NH4Cl: 38.944 g
(i) Mass of original sample = Mass of evaporating dish and sample after subliming NH4Cl - Mass of evaporating dish #1 Mass of original sample
= 38.944 g - 38.120 g
= 0.824 g
(ii) Mass of NH4Cl(g)
= Mass of evaporating dish and original sample - Mass of evaporating dish and sample after subliming NH4Cl Mass of NH4Cl(g)
= 39.070 g - 38.944 g
= 0.126 g
(iii) Percent by mass of NH4Cl(%)
= (mass of NH4Cl(g) / Mass of original sample) x 100% Percent by mass of NH4Cl(%)
= (0.126 g / 0.824 g) x 100%
= 15.291%
(iv) Mass of evaporating dish #2: Not given Mass of watch glass: Not given Mass of evaporating dish #2, watch glass, and NaCl: Not given
(v) Mass of NaCl(g)
= Mass of evaporating dish and NaCl - Mass of evaporating dish #2 - Mass of watch glass Mass of NaCl(g)
= (38.360 g - 38.120 g) - 20.000 g
= 0.240 g
(vi) Percent by mass of NaCl(%)
= (mass of NaCl(g) / Mass of original sample) x 100% Percent by mass of NaCl(%)
= (0.240 g / 0.824 g) x 100%
= 29.126%.
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Suppose 8.49 g of sodium bromide is dissolved in 200, mL of a 0.50M aqueous solution of silver nitrate. Calculste the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium bromide is dissolved in it. Be sure your answer has the correct number of significant digits:
There are no Na+ ions left in solution after the reaction is complete, the final molarity of Na+ ions in the solution is 0 M. The final molarity of sodium cation in the solution is 0 M.
In order to calculate the final molarity of sodium cation in the solution, first, we need to calculate the number of moles of sodium bromide. We can use the formula for the number of moles, given as:
Number of moles = Mass / Molar mass
The molar mass of sodium bromide (NaBr) is 102.89 g/mol.Number of moles of
NaBr = 8.49 g / 102.89 g/mol= 0.0825 mol
Now, we have to calculate the number of moles of silver nitrate (AgNO3) in 200 mL of 0.50 M aqueous solution.
Since we are given the volume in mL, we need to convert it into liters (L) first:
1 L = 1000 mL
So, the volume in liters is 200/1000 = 0.2 L
The formula for the number of moles is:
Number of moles
= Molarity x Volume (in liters)Number of moles of AgNO3
= 0.50 M x 0.2 L
= 0.1 mol
Now, we have to find out the limiting reactant.
This is because one of the reactants will be consumed completely and the other will be left in excess.
The reactant that gets completely consumed is the limiting reactant, and the number of moles of the product (in this case, sodium cation) depends on it.
So, we need to compare the number of moles of NaBr and AgNO3:Number of moles of NaBr = 0.0825 mol
Number of moles of AgNO3 = 0.1 mol
From the above comparison, we can see that AgNO3 is the limiting reactant since the number of moles of NaBr is less than the number of moles of AgNO3.
Therefore, all of the Na+ ions will react with NO3- ions to form AgBr, and there will be no Na+ ions left in solution.
Finally, we can calculate the final molarity of Na+ ions in solution.
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Answer: To calculate the final molarity of sodium cation (Na+) in the solution
Explanation:
Number of moles of NaBr:
Molar mass of NaBr = 22.99 g/mol (Na) + 79.90 g/mol (Br) ≈ 102.89 g/mol
Number of moles of NaBr = 8.49 g / 102.89 g/mol ≈ 0.0825 mol (rounded to 4 significant digits)
Number of moles of AgNO3 required to react with NaBr:
The balanced chemical equation for the reaction is:
NaBr + AgNO3 → NaNO3 + AgBr
From the equation, we can see that 1 mole of NaBr reacts with 1 mole of AgNO3.
Thus, the number of moles of AgNO3 required = 0.0825 mol (rounded to 4 significant digits)
Number of moles of Na+ = 0.0825 mol (rounded to 4 significant digits)
Next, we need to calculate the total volume of the solution after the reaction, which will be the same as the initial volume since the volume doesn't change when the sodium bromide dissolves in it. The total volume is 200 mL.
Now, let's calculate the final molarity of sodium cation (Na+):
Molarity (M) = moles of solute / volume of solution (in liters)
Volume of solution in liters = 200 mL = 200 mL / 1000 mL/L = 0.2 L
Final molarity of Na+ = 0.0825 mol / 0.2 L = 0.4125 M
So, the final molarity of sodium cation (Na+) in the solution is approximately 0.4125 M, rounded to 4 significant digits.
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suppose you place 3.55 ml of a substance into a graduated cylinder. the graduated cylinder has a mass of 12.55 g when empty and a mass of 15.08 g after adding the substance.
The mass of the substance is calculated by subtracting the empty graduated cylinder's mass from the mass of the cylinder with the substance: 15.08 g - 12.55 g = 2.53 g.
We subtract the empty graduated cylinder's mass from the substance's mass to find its mass. The empty cylinder weighs 12.55 g, whereas the filled one weighs 15.08 g. 2.53 g separates them.
The graded cylinder's substance added 2.53 g. This calculation assumes the graduated cylinder does not affect measured mass.
The substance's mass (2.53 g) is calculated by subtracting the initial (12.55 g) from the final (15.08 g). This method uses the mass difference before and after adding a substance to a container to measure its mass.
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n the following experiment a coffee-cup calonmeter contaning 100.0 mL of H 2
O is used. The inital temperature of the calonmeter is 230 C. if to 0 g of CaCh a added to the calorimeter, what wil be the linal temperature of the solution in the calorimeter? The enthapy of dissolution Δ. H of CaCla 5−82.8k I mol 1 . Express your answer with the appropriate units.
The final temperature of the solution in the calorimeter is 22.65°C.
A coffee-cup calorimeter containing 100.0 mL of H2O is used in the given experiment. The initial temperature of the calorimeter is 23°C. If 0.20 g of CaCl2 is added to the calorimeter,
The enthalpy of dissolution (ΔH) of CaCl2 = -82.8 kJ/mol.
To determine the final temperature of the solution in the calorimeter, we will use the following formula:Q = m × c × ΔTWhere Q is the amount of heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the temperature change.
The mass of the solution is calculated by taking the density of water (1 g/mL) and multiplying it by the volume of the solution (100 mL):
m = 1.00 g/mL × 100.0 mL
= 100.0 g
The specific heat capacity of water is 4.18 J/g°C, so:
c = 4.18 J/g°C
The temperature change can be calculated as follows:
ΔT = Q / (m × c)
The amount of heat transferred can be found using the enthalpy of dissolution of
CaCl2:ΔH = -82.8 kJ/mol
The number of moles of CaCl2 added to the calorimeter can be calculated as follows:
n = m / M
where M is the molar mass of CaCl2:M = 110.98 g/moln = 0.20 g / 110.98 g/moln = 0.00180 molThe amount of heat transferred can be calculated as follows:
Q = n × ΔHQ = (0.00180 mol) × (-82.8 kJ/mol)Q
= -0.149 kJ = -149 J
Finally, we can use the formula above to calculate the temperature change:
ΔT = Q / (m × c)ΔT
= (-149 J) / (100.0 g × 4.18 J/g°C)ΔT
= -0.355°C
So the final temperature of the solution in the calorimeter is 23°C - 0.355°C = 22.65°C.
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when a piece of metal was heated in a flame and then dropped into 2.00 x 102 ml of water at 22.5°c, the temperature of the water rose to 38.7°c. how much heat was transferred from the metal to the water?
The amount of heat transferred from the metal to the water is approximately 134,064 Joules.
To calculate the amount of heat transferred from the metal to the water, you can use the formula:
Q = m × c × ΔT
Where:
Q is the heat transferred (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (approximately 4.18 J/g°C)
ΔT is the change in temperature (in °C)
First, you need to determine the mass of the water. The volume of the water is given as 2.00 x 10² mL, which is equivalent to 2.00 x 10² g (since the density of water is approximately 1 g/mL).
Next, calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 38.7°C - 22.5°C
Now, you can calculate the amount of heat transferred:
Q = m × c × ΔT
Substituting the values:
Q = (2.00 x 10²g) × (4.18 J/g°C) × (38.7°C - 22.5°C)
Calculate the value to find the amount of heat transferred from the metal to the water in Joules.
Q = (2.00 x 10²) × (4.18) × (16.2)
Calculating the final value:
Q ≈ 134,064 Joules
Therefore, the amount of heat transferred from the metal to the water is approximately 134,064 Joules.
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You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100M benzoic acid (pK
a
=4.20) and 0.240M sodium benzoatc. How many milliliters of each solution should be mixed to prepare this buffer? benzoic acid:
Previous question
To prepare the pH 4.00 buffer solution, you should mix approximately 61.35 mL of the 0.100 M benzoic acid solution with 38.65 mL of the 0.240 M sodium benzoate solution.The ratio of benzoic acid to sodium benzoate in the buffer solution using the Henderson-Hasselbalch equation.
To prepare a pH 4.00 buffer solution using benzoic acid and sodium benzoate, we need to calculate the appropriate volumes of the 0.100 M benzoic acid and 0.240 M sodium benzoate solutions.
First, we need to determine the ratio of benzoic acid to sodium benzoate in the buffer solution. The Henderson-Hasselbalch equation can help us with this calculation:
pH = pKa + log([A-]/[HA])
Given that the pH is 4.00 and pKa is 4.20, we can rearrange the equation:
log([A-]/[HA]) = pH - pKa
log([A-]/[HA]) = 4.00 - 4.20
log([A-]/[HA]) = -0.20
Next, we take the antilog of -0.20 to find the ratio of [A-] to [HA]:
[A-]/[HA] = antilog(-0.20)
[A-]/[HA] = 0.63
The ratio of [A-] to [HA] is 0.63.
Now, let's calculate the volumes of each solution needed. Let's assume x represents the volume (in mL) of the 0.100 M benzoic acid solution and y represents the volume (in mL) of the 0.240 M sodium benzoate solution.
Since the total volume is 100.0 mL, we have the equation: x + y = 100
Considering the ratio of [A-] to [HA] as 0.63, we can write the equation: y/x = 0.63
Solving these two equations simultaneously will give us the volumes of each solution:
x + y = 100
y/x = 0.63
By substituting y = 0.63x from the second equation into the first equation, we get:
x + 0.63x = 100
1.63x = 100
x = 61.35 mL (rounded to two decimal places)
Substituting this value back into the equation x + y = 100, we find:
61.35 + y = 100
y = 38.65 mL (rounded to two decimal places)
Therefore, to prepare the pH 4.00 buffer solution, you should mix approximately 61.35 mL of the 0.100 M benzoic acid solution with 38.65 mL of the 0.240 M sodium benzoate solution.
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What is the tonicity of the solution made of 0.3m glucose after the blood is added when the membrane is impermeable to glucose and permeable to water? glucose is a covalent molecule.
The tonicity of a solution refers to its ability to cause a change in the shape or size of cells by altering the water content.
In this case, since the membrane is impermeable to glucose but permeable to water, the glucose molecules cannot pass through the membrane, while water molecules can. Since the solution is made of 0.3M glucose, it means that the concentration of glucose in the solution is 0.3 moles per liter.
When blood is added, the impermeable membrane prevents glucose molecules from passing through, but water molecules can move freely. The presence of a higher concentration of glucose inside the membrane than in the blood creates a hypertonic environment. This causes water to move from the blood (where the concentration of solutes is lower) into the solution, via osmosis.
As a result, the solution will become more diluted as water enters it, causing it to expand and potentially change the shape or size of the cells.
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Enter your answer in the provided box. Carry out the following calculation, making sure that your answer has the correct number of significant figures: 2.210 cm+12.5 cm+176.0 cm+318 cm=cm
We are adding all the numbers together, we must make sure that our answer has the same number of significant figures as the number with the least significant figures in the addition,
which is two.
To calculate the value of
2.210 cm + 12.5 cm + 176.0 cm + 318 cm,
we can add the numbers together as shown below:
2.210 cm+12.5 cm+176.0 cm+318 cm
= 508.71 cm (rounded to two significant figures)
Therefore, the sum of 2.210 cm, 12.5 cm, 176.0 cm and 318 cm is 508.71 cm, rounded to two significant figures.
Note that we rounded the answer to two significant figures because 2.210 cm has only three significant figures.
We are adding all the numbers together, we must make sure that our answer has the same number of significant figures as the number with the least significant figures in the addition, which is two.
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The ph of a 0.21 msolution of 3-hydroxypropanoic acid is measured to be .calculate the acid dissociation constant of 3-hydroxypropanoic acid. round your answer to significant digits.
The acid dissociation constant (Ka) of 3-hydroxypropanoic acid is approximately 0.309, rounded to significant digits.
To calculate the acid dissociation constant (Ka) of 3-hydroxypropanoic acid, we need to use the measured pH and the concentration of the acid solution.
Given:
pH of the 0.21 M solution = (to be determined)
Concentration of 3-hydroxypropanoic acid = 0.21 M
To calculate Ka, we need to consider the dissociation of the acid into its conjugate base and hydrogen ions:
3-hydroxypropanoic acid ⇌ 3-hydroxypropanoate⁻ + H⁺
The dissociation of the acid can be represented by the equation:
Ka = [3-hydroxypropanoate⁻] × [H⁺] / [3-hydroxypropanoic acid]
Since the concentration of the acid and its conjugate base are the same initially, and we assume complete dissociation, we can simplify the equation to:
Ka = [H⁺]² / [3-hydroxypropanoic acid]
To find [H⁺], we can use the pH value:
[H⁺] = [tex]10^(-pH)[/tex]
Substituting the given values:
[H⁺] = [tex]10^(-pH) = 10^(-0.21)[/tex]
Now, we can substitute the values into the Ka equation:
Ka = [H⁺]² / [3-hydroxypropanoic acid] =[tex](10^(-0.21))² / 0.21[/tex]
Using a calculator:
Ka ≈ 0.309
Therefore, the acid dissociation constant (Ka) of 3-hydroxypropanoic acid is approximately 0.309, rounded to significant digits.
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9. Which of the following neutral atoms has the largest first ionization energy? Ne p Zn Cl k 10. Calculate △E for a system that releases 41 J of heat while 28 J of work is done by the system. 41 J 13 J −13 J 69 J −69 J
The given values are:q = -41 J (negative sign indicates that the heat is released by the system)w = 28 JΔE = q + w= (-41 J) + 28 J= -13 J
Therefore, the answer is -13 J.
The element that has the largest first ionization energy among Ne, P, Zn, Cl, and K is Ne.
The first ionization energy (IE1) is defined as the amount of energy required to remove one mole of an electron from one mole of a gaseous element to form one mole of gaseous cation with a positive charge of 1.
The first ionization energy of an atom is determined by the nuclear charge and the atomic radius.
The nuclear charge is the number of protons in the nucleus, which determines the number of electrons, and the atomic radius is the distance between the nucleus and the outermost shell where the valence electrons are located.
The element with the highest first ionization energy would have a high nuclear charge and a small atomic radius. Among the given elements, the element that satisfies this condition is neon (Ne).
Therefore, the answer is Ne.10.
The formula for the calculation of ΔE is:ΔE
= q + w
where ΔE represents the change in internal energy of a system, q is the heat absorbed or released by the system, and w is the work done on or by the system.
The given values are:q
= -41 J (negative sign indicates that the heat is released by the system)w
= 28 JΔE
= q + w
= (-41 J) + 28 J
= -13 J
Therefore, the answer is -13 J.
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acetic acid can be deprotonated to form acetate ion, ch3co–2 . draw two other resonance structures. identify the major (most important) and minor (less important) structures.
There are two additional resonance structures that can be drawn when acetic acid (CHCOOH) is deprotonated to yield the acetate ion [tex](CH_3COO^-).[/tex] The resonance structures are listed below:
1. [tex]CH_3COO^-[/tex]is the major resonance structure (most important).
In this form the oxygen atom has a negative charge, indicating that the extra electron from the precipitate is concentrated on it.
2. Minor Resonance Structure (Less Important):
[tex]CH_2=CO-O-[/tex]
In this structure, the double bond moves to the carbon–oxygen bond, leaving the oxygen atom with a negative charge and the next carbon atom with a positive charge.
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What is the molality of a 2 M solution of sodium acetate in
water with a density of 1.23 g/mL?
Molality is a measure of concentration that denotes the number of moles of solute per kilogram of solvent. Therefore, the molality of a 2 M solution of sodium acetate in water with a density of 1.23 g/mL is 1.63 m.
Molality is calculated by the formula :
molality (m) = moles of solute / mass of solvent (in kilograms)
The mass of solvent in kilograms can be calculated using the density of the solution.
The formula is:
Mass = density x volume.
To calculate the molality of a 2 M solution of sodium acetate in water with a density of 1.23 g/mL, we will need to use the above formulas.
Here's how:
First, we need to determine the mass of 1 L of the solution:
mass = density x volume
mass = 1.23 g/mL x 1000 mL
mass = 1230 g
Next, we need to convert the mass to kilograms:
mass = 1230 g x (1 kg/1000 g)
mass = 1.23 kg
We also need to determine the number of moles of sodium acetate in 1 L of solution.
To calculate the moles, we use the molarity (M) and the volume (V) of the solution:
moles = M x V
moles = 2 M x 1 L
moles = 2 moles
Finally, we can use these values to calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
m = 2 moles / 1.23 kg
m = 1.63 m
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You need to make 250 ml of a stock solution of 0.1 m na2 atp. its formula weight is 605.2 g mol-1 . how much na atp should you weigh out?
Weigh out approximately 15.13 grams of Na2ATP to make a 250 ml stock solution of 0.1 M Na2ATP.
To make a 250 ml stock solution of 0.1 M Na2ATP, you need to calculate the amount of Na2ATP in grams.
First, determine the number of moles required using the formula:
moles = Molarity x Volume (in liters) moles = 0.1 M x 0.250 L
moles = 0.025 mol
Next, calculate the mass of Na2ATP using the formula:
mass = moles x formula weight mass = 0.025 mol x 605.2 g/mol mass = 15.13 g
Therefore, you should weigh out approximately 15.13 grams of Na2ATP to make a 250 ml stock solution of 0.1 M Na2ATP.
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which of the following accurately describes the ph scale? which of the following accurately describes the ph scale? the ph scale runs from 0 (neutral) to 14 (most acidic), with 7 as an average acidity level. the ph scale runs from 0 (most acidic) to 14 (neutral), with 7 as an average acidity level. the ph scale runs from 0 (most basic) to 14 (most acidic), with 7 as a neutral. the ph scale runs from 0 (most acidic) to 14 (most basic), with 7 as a neutral.
Answer:
The pH scale measures acidity of a substance. known as potential of hydrogen, it varies from 0 to 14 with 7 being the pH value of a neutral solution. Below 7 shows the substance is acidic in nature and above 7 is alkaline in nature. pH 0-3 are considered strong acids while pH 4-6 are weak acids. pH 8-10 are weak alkalines and pH 11-14 are strong alkalines. This is a general trend and there may be exeptions especially if the substance has a negative pH. However, it would not be covered likely unless you are doing university chemistry.
What is the rate constant of a first-order reaction that takes 354 seconds for the reactant concentration to drop to half of its initial value?
The rate constant of a first-order reaction can be calculated using the formula k = ln(2) / t, where k is the rate constant and t is the time it takes for the reactant concentration to drop to half of its initial value.
In this case, the time given is 354 seconds. Using the formula, we can calculate the rate constant:
k = ln(2) / 354
k ≈ 0.00196 s^-1
The rate constant of a first-order reaction represents the speed at which the reaction occurs. It is specific to each reaction and is independent of the initial concentration of the reactant. In this case, the rate constant is approximately 0.00196 s^-1.
The rate constant of a first-order reaction is an important parameter in chemical kinetics. It determines the rate at which the reaction proceeds.
In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. As the reactant concentration decreases, the rate of reaction decreases. The rate constant is calculated by using the natural logarithm of 2 divided by the time it takes for the reactant concentration to halve. In this case, the given time is 354 seconds. Plugging this value into the formula, the rate constant is approximately 0.00196 s^-1. This means that the reaction proceeds at a rate of 0.00196 units per second. The rate constant is a characteristic of the specific reaction and can be used to determine the reaction kinetics and predict the reaction's behavior under different conditions.
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How does a homogeneous mixture differ from a pure substance?
Considering the definition of pure substance and homogeneous mixture, the main difference is that a pure substance consists of only one type of particle and it cannot be separated or divided into more substances whereas a homogeneous mixture is made up of two or more different substances and can be separated into various pure substances.
Definition of pure substance and homogeneous mixtureA pure substance is one that is made up of a single type of particle, whether atoms or molecules, and therefore has the same properties in all its parts. The composition and properties of an element or compound are uniform anywhere in a given sample, or in different samples of the same element or compound.
When a substance is made up of two or more simple substances, it is known as a mixture. Homogeneous mixtures are characterized by being formed by two or more components that cannot be distinguished visually. The composition and properties are uniform throughout any given sample, but may vary from sample to sample. In general, the components of a homogeneous mixture can be in any proportion, and can be recovered using physical separation methods.
DifferenceThe main difference between a pure substance and a mixture is that a pure substance consists of only one type of particle and it cannot be separated or divided into more substances whereas a homogeneous mixture is made up of two or more different substances and can be separated into various pure substances.
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what is buckminsterfullerene in chemistry
Buckminsterfullerene, also known as C60 or buckyball, is a unique and fascinating molecule in the field of chemistry.
It was first discovered in 1985 by a team of scientists led by Richard Smalley, Robert Curl, and Harold Kroto, who were awarded the Nobel Prize in Chemistry in 1996 for their discovery.
Buckminsterfullerene is a carbon allotrope composed of 60 carbon atoms arranged in a hollow sphere resembling a soccer ball. Its name is derived from its resemblance to the geodesic dome designs created by architect Buckminster Fuller.
One of the remarkable aspects of buckminsterfullerene is its symmetrical structure, which confers extraordinary stability. Its structure allows for the distribution of strain throughout the molecule, making it highly resistant to chemical reactions and providing exceptional thermal and mechanical stability.
Buckminsterfullerene exhibits a range of unique properties that have attracted significant scientific interest. It is an excellent electron acceptor and can undergo various chemical reactions due to its high reactivity. Its electronic properties have applications in organic electronics, photovoltaics, and molecular electronics.
Moreover, buckminsterfullerene has shown potential in various fields, including medicine, material science, and nanotechnology. Its hollow structure can encapsulate other atoms or molecules, making it useful for drug delivery systems.
In summary, buckminsterfullerene is a fascinating carbon molecule with a distinctive structure and exceptional properties. Its discovery has opened up new avenues for research and applications in chemistry, physics, materials science, and other interdisciplinary fields.
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Which electron orbital diagram is written correctly for an atom without any violations?
An atom without any violations will have all of its electrons placed in the lowest energy levels and is in the ground state. The correct electron orbital diagram for an atom without any violations is the one that adheres to the rules regarding the filling of electrons in the orbitals.
An electron orbital diagram is a representation of an atom in which the atomic nucleus is shown in the center and the electrons are represented in their appropriate orbitals. The electron configuration of an atom can be represented in an electron orbital diagram.
The following rules should be considered while drawing electron orbital diagrams:
There are four different types of orbitals: s, p, d, and f. s orbitals hold a maximum of two electrons, p orbitals hold a maximum of six electrons, d orbitals hold a maximum of ten electrons, and f orbitals hold a maximum of fourteen electrons. The orbital with the lowest energy level is the first to be filled.
According to the Aufbau Principle, the lower energy level orbitals must be filled before the higher energy level orbitals. Each orbital must be filled with one electron before any orbital can be filled with a second electron.
Electrons in orbitals of the same energy must be present before electrons in orbitals of higher energy can be present. For atoms in the ground state, electrons must be placed in the lowest energy level orbitals before they can be placed in higher energy level orbitals.
So, the correct electron orbital diagram for an atom without any violations is the one that adheres to these rules regarding the filling of electrons in the orbitals.
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