The reaction between calcium nitrate (Ca(NO₃)₂) and sodium oxalate (Na₂C₂O₄) can be represented by the following chemical equation:
Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2NaNO₃
This is a double displacement reaction, where the calcium ion (Ca²⁺) from calcium nitrate and the oxalate ion (C₂O₄²⁻) from sodium oxalate switch places to form calcium oxalate (CaC₂O₄) and sodium nitrate (NaNO₃). The balanced chemical equation shows that one mole of calcium nitrate reacts with one mole of sodium oxalate to form one mole of calcium oxalate and two moles of sodium nitrate.
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We've figured out what part of the salt causes the flame to change color, so now let's measure the wavelengths created with four metals.
Use the ruler under the "tools" icon in the upper right of the video player to measure the wavelengths of light released by each compound.
Around 450 nm is the wavelength of the spectral line for potassium chloride. The distance among identical spots between two succeeding waves is known as the wavelength.
The distance among identical spots between two succeeding waves is known as the wavelength, which is a feature of waves. The wavelength of a wave is the distance across one wave's peak (or trough) and the next. In mathematics, the Greek symbol lambda () is used to denote wavelength. The colour of light is determined by its wavelength, and the pitch of sound is determined by its wavelength. Around 450 nm is the wavelength of the spectral line for potassium chloride.
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(25 pts) Explain in depth the relationship between LIMITING reactant and THEORETICAL yield
Answer: I got you fam
Explanation:
A limiting reactant is a reactant in a chemical reaction that limits the amount of product created.
So for example if there are elements X and Y reacting to create product XY, once say element X runs out, the reaction stops, even though there is still more of the reactant Y. So there is 0 g of element X remaining, and maybe 2 g left of element Y. X is the limiting reactant since it limits the amount of product made.
Theoretical yield is the maximum amount of product that could be made in an experiment. This occurs if a reaction is 100% effective (and experimentally, this doesn't usually happen, which is why it is called theoretical).
Write short note on the
-physical and chemical methods
of monitoring the rate of
chemical reaction
Answer:
Physical and chemical methods can be used to monitor the rate of a chemical reaction. Physical methods measure changes in properties like temperature, pressure, or volume. Chemical methods track reactant consumption or product formation using techniques like titration or spectrophotometry. The choice of method depends on the reaction being studied, and scientists use these methods to gain insight into reaction kinetics and optimize conditions for better efficiency and selectivity.
6. Calculate the pH of a buffer prepared by mixing 0.10 mol-L-' acetic acid (CH;COOH, HAc) and
0.10 mol L NaOH solution with the volume ratio of 3:1. (Answer pH = 4.45
in diagram A, what is the value of ^H
The value of h in the right triangle is 7.7 cm.
Why is this so?Trigonometric ratio is used to show the relationship between the sides and angles of a right angled triangle.
In the right triangle, using trigonometric ratio:
a) sin(31) = h/15
⇒ 0.515 = h/15
Making H the subject of the formula we have:
0.515 x 15 = h/15 x 15
h = 7.725
h [tex]\approx[/tex] 7,7
Hence, is is correct to state that the value of h in the diagram is 7.7 cm approximately. See the attached image.
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Full Question:
Although part of your question is missing, you might be referring to this full question:
the diagram shows a right-angled triangle. what is the value of h? give your answer correct to 1 decimal place.
How many liters of a 3.4 m isopropanol solution can be made with 78 g of isopropanol
Answer:
You can make 2.29 L of 3.4 M isopropanol solution with 78 g of isopropanol.
Explanation:
A student's sample of Alka Seltzer (0.350 g) contained 0.207 g of sodium bicarbonate (NaHCO3). How many moles of NaHCO3 are in their sample?
Group of answer choices
The number of moles of the sodium bicarbonate is 0.0025 moles
What is the number of moles?We know that the moles is the amount of the substance that is contained and that we can get the moles as the ratio of the mass to the molar mass of the substance and that is what we are going to do in the case of the sodium bicarbonate that we have in the question that is here.
Now we have from the question that;
Mass of the sodium bicarbonate = 0.207 g
Number of moles of the sodium bicarbonate can now be gotten in this case by the use of the formula;
Number of moles
0.207 g/84 g/mol
= 0.0025 moles
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1b. Suppose that you were titrating a 100 mL acid solution with the 0.1 M NaOH solution that you made. You performed the titration multiple times and obtained the data below. Complete the data table below. Show work on a separate piece of paper/ the back of this paper.
Step 1: Write and balance the chemical equation (only need to do this once for each titration)
Step 2: Use the molarity and mL of base used to find the moles of base it took to neutralize the acid
Step 3: Calculate moles of acid neutralized
Step 4: Calculate molarity of acid
Step 5: Calculate pH
1c. Calculate the most likely pH of the acid solution by finding the average of all the pH's you found in each of your multiple titrations. We find the average to minimize human errors made while titrating.
The moles of NaOH used is 0.0008 moles
The molarity of the acid is 0.008 M
What is the molarity of the acid?The molarity of the acid is found as follows:
Moles of NaOH used = concentration of NaOH × volume of NaOH used
the average volume of NaOH used = 8.0 mL
moles of NaOH = 0.1 M × 8.0 mL
moles of NaOH = 0.0008 moles
Molarity of acid:
Assuming the acid is monobasic, the mole ratio of acid to base is 1 : 1
The volume of acid used is 100 mL
The molarity of acid = moles of acid / volume of acid in liters
The molarity of acid = 0.0008 moles / 0.1 L
The molarity of acid = 0.008 M
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A piece of metal with a mass of 32.8 g is heated to 100.5 C and dropped into 138.2 g of water at 20.0 C. the final temperature of the system is 30.2 C. What is the specific heat capacity of the metal
Answer:
To solve this problem, we can use the equation:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
In this case, we know that the heat lost by the metal is equal to the heat gained by the water:
Q lost = Q gained
We can calculate the heat lost by the metal using the equation:
Q lost = m * c * ΔT
where m is the mass of the metal, c is the specific heat capacity of the metal (which we are trying to find), and ΔT is the change in temperature of the metal (100.5 C - 30.2 C = 70.3 C).
We can calculate the heat gained by the water using the equation:
Q gained = m * c * ΔT
where m is the mass of the water and ΔT is the change in temperature of the water (30.2 C - 20.0 C = 10.2 C).
Setting the two equations equal to each other, we get:
m * c * ΔT (metal) = m * c * ΔT (water)
Simplifying, we get:
c (metal) = (m * c * ΔT (water)) / (m * ΔT (metal))
Plugging in the values we know:
m (metal) = 32.8 g
ΔT (metal) = 70.3 C
m (water) = 138.2 g
ΔT (water) = 10.2 C
c (metal) = (138.2 g * 4.184 J/g·K * 10.2 C) / (32.8 g * 70.3 C)
c (metal) = 0.192 J/g·K
Therefore, the specific heat capacity of the metal is 0.192 J/g·K.
20. Using the equilibrium expression from Question #19 and based upon the value of K, would you expect there to be more or less product in the equilibrium mixture? Explain why.
The value of K is given as 2344 thus more products are obtained at equilibrium since the equilibrium constant is very large.
The equilibrium constant ( K ) is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
The balanced equation is -
CaO(s) + CH₄(g) + 2H₂O(g) —> CaCO₃(s) + 4H₂(g)
K = [CaCO₃] [H₂]⁴ / [CaO] [CH₄] [H₂O]²
Since the value of K = 2344, which is large, more of products will be formed in the reaction.
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Calculate the percentage by mass of oxygen in copper (II) tetraoxosulphate (VI) (Cu = 64.0, S = 32.0, 0 = 16.0)
The percentage by mass of oxygen in CuSO4 is 40%.
What is the percentage by mass of an atom in a compound?Finding the compound's molar mass is the first step in calculating the percentage of an atom by mass. The atomic masses of all the atoms in the compound are added up to determine the molar mass, which is the mass of one mole of the molecule.
We have the molar mass of the CuSO4 is given as;
64 + 32 + 4(16) = 160 g/mol
Then we have the percentage by mass of oxygen as;
4(16)/160 * 100/1
40%
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Substance A decomposes at a rate proportional to the amount of A present.
Write an equation relating A to the amount left of an initial amount Ao after time t.
It is found that 8 g of A will reduce to 4 g in 3 hr. After how long will there be only 1 g left?
It will take approximately 2.079 hours for there to be only 1 g of substance A left.
The rate of decomposition of substance A is proportional to the amount of A present, which means that we can use the following differential equation to describe the decay;
dA/dt = -kA
where A is the amount of substance A at time t, k is the rate constant of the reaction, and the negative sign indicates that A is decreasing over time.
To solve differential equation, we separate the variables and integrate;
[tex]d_{A}[/tex]/A = -k [tex]d_{t}[/tex]
Integrating both sides gives;
ln(A) = [tex]-k_{t}[/tex] + C
where C is the constant of integration. To find the value of C, we can use the initial condition that 8 g of A reduces to 4 g in 3 hours. At t=0, A=Ao=8 g, and at t=3, A=4 g. Substituting values into the equation above, we have;
ln(8) = -3k + C
ln(4) = -6k + C
Subtracting first equation from second, we get;
ln(4/8) = -3k
Simplifying, we get;
k = ln(2)/3
Now, we can use the equation we derived earlier to find how long it will take for there to be only 1 g of A left;
ln(A) =[tex]-k_{t}[/tex] + C
ln(1) = -(ln(2)/3)t + C
Simplifying, we get:
t = 3 ln(2)
t ≈ 2.079 hours
Therefore, it will take 2.079 hours.
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We would like to find RDS using Tapel slope. Provide Tafel slope when we assume each step is RDS, alpha a=0.5 * Target Reaction : Cu oxidation [mV]
Mechanism1 Cu-> Cu2+ +2e-
Mechanism2 Cu-> Cu+ +e-
Cu+->Cu2+ +e-
The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::
b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V
How to solveTo identify the rate-controlling step (RCS) utilizing the Tafel slope, we initially need to calculate the Tafel slope for each suggested mechanism when the electron transfer coefficient (alpha, α) equals 0.5.
The target reaction involves Cu oxidation.
Mechanism 1:
Cu -> Cu²⁺ + 2e⁻
Mechanism 2:
Cu -> Cu⁺ + e⁻
Cu⁺ -> Cu²⁺ + e⁻
The Tafel slope (b) can be computed with the following formula:
b = (2.303 * R * T) / (α * n * F)
Where:
R signifies the gas constant (8.314 J/mol K)
T represents the temperature in Kelvin (let's assume 298 K, standard room temperature)
α denotes the electron transfer coefficient (0.5)
n is the number of electrons exchanged in the RCS
F is the Faraday constant (96,485 C/mol)
For Mechanism 1, n = 2 (since 2 electrons are exchanged in the rate-controlling step):
b1 = (2.303 * 8.314 * 298) / (0.5 * 2 * 96,485) = 0.0592 V
For Mechanism 2, we must examine both steps. Let's initially evaluate the Tafel slope for each step.
Step 1 (n = 1):
b2_1 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V
Step 2 (n = 1):
b2_2 = (2.303 * 8.314 * 298) / (0.5 * 1 * 96,485) = 0.1184 V
The Tafel slope for Mechanism 2 is the sum of the slopes of both steps, presuming that each step in Mechanism 2 is RCS::
b2 = b2_1 + b2_2 = 0.1184 + 0.1184 = 0.2368 V
Having obtained the Tafel slopes for both mechanisms, we can now compare them to the experimental Tafel slope to ascertain which mechanism is more likely the RCS.
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The symbol that indicates a substance dissolve in water is
The symbol that indicates a substance dissolve in water is (aq).
What is the symbol?The word "(aq)" stands for aqueous and denotes a material that dissolves in water.
Aqueous solutions are created when substances dissolve in water and are uniformly dispersed throughout the liquid. "(aq)" is added to the end of a substance's chemical formula in a chemical equation to denote that the material is in an aqueous state.
We often see this in several chemical reactions and the symbols shows that the solute was dissolved in water.
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How much energy is required to heat 186 grams of water from 32 degrees Celsius to 90
degrees Celsius?
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Reaction Ai Sodium Bicarbonate and Hydrochloric Acid
Experimental Data
(a) Mass of evaporating dish
watch glass
(b) Mass of evaporating dish watch glass sodium bicarbonate
(c) Mass of sodium bicarbonate used
(d) Mass of evaporating dish watch glass sodium chloride
(e) Mass of sodium chloride collected (experimental yield)
.
Mole Ration and Reaction Story
.
.
Data Analysis
1) Use your data to determine the experimental mole-to-mole ratio between sodium bicarbonate and sodium chloride
Show your work for each
NaHCOS
Convert the mass of sodium bicarbonate used to moles
100.69
1
mole
9
Convert the mass of sodium chloride collected in moder
Nac
2g
104.2
3.bg
mole
g
Divide both of your results from the preceding two steps by the lower mole value to determine the simplest mole-to-
mole ratio between sodium bicarbonate and sodium chloride.
The reaction between sodium carbonate and hydrochloric acid
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
How to solveFor reaction A
Mass of sodium bicarbonate used = (Mas of evaporating dish + watch glas + sodium bicarbonate) - (Mas of evaporating dish + watch glas)
= 46.582 - 46.263
= 0.319 g
Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)
= 46.473 - 46.263
= 0.210 g
Moles of sodium bicarbonate (NaHCO3) used = mas/molecular weight
= (0.319 g) / (84 g/mol)
= 0.00380 mol
Moles of sodium chloride (NaCl) used = mas/molecular weight
= (0.210 g) / (58.44 g/mol)
= 0.00359 mol
Mol ratio of NaHCO3 : NaCl = 0.00380 : 0.00359
Divide by 0.00359
Simple mol ratio
NaHCO3 : NaCl = 1.06 : 1
After rounding
Mol ratio of NaHCO3 : NaCl = 1 : 1
Moles of NaHCO3 = moles of NaCl = 0.00359 mol
Theoretical yield of NaCl = moles x molecular weight
= 0.00359 mol x 58.44 g/mol
= 0.210 g
the percent yield of sodium chloride
= actual yield x 100 / theoretical yield
= 0.210*100/0.210
= 100%
the reaction between sodium bicarbonate and hydrochloric acid
NaHCO3 + HCl = NaCl + CO2 + H2O
For reaction B
Mass of sodium carbonate used = (Mas of evaporating dish + watch glas + sodium carbonate) - (Mas of evaporating dish + watch glas)
= 51.677 - 51.368
= 0.309 g
Mass of sodium chloride = (mas of evaporating dish + watch glas + sodium chloride) - (Mas of evaporating dish + watch glas)
= 51.671 - 51.368
= 0.303 g
Moles of sodium carbonate (Na2CO3) used = mas/molecular weight
= (0.309 g) / (106 g/mol)
= 0.00292 mol
Moles of sodium chloride (NaCl) used = mas/molecular weight
= (0.303 g) / (58.44 g/mol)
= 0.00518 mol
Mol ratio of
Na2CO3 : NaCl = 0.00292 : 0.00518
Divide by 0.00292
Simple mol ratio
Na2CO3 : NaCl = 1 : 1.78
After rounding
Mol ratio of Na2CO3 : NaCl = 1 : 2
Moles of NaCl = 2 x moles of Na2CO3
= 2 x 0.00292 = 0.00584 mol
Theoretical yield of NaCl = moles x molecular weight
= 0.00584 mol x 58.44 g/mol
= 0.341 g
the percent yield of sodium chloride
= actual yield x 100 / theoretical yield
= 0.303*100/0.341
= 88.86%
the reaction between sodium carbonate and hydrochloric acid
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
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find the ΔG°rxn for a reaction that has a K = 17.23 at 371 K. Is the reaction spontaneous. Note: R = 8.314 J/mol.K
The value of ΔG°rxn is - 6979 J/mol, and the reaction is spontaneous.
The change in Gibbs free energy (ΔG°rxn) for a reaction can be calculated using the equation;
ΔG°rxn = -RT ln(K)
where; R = gas constant = 8.314 J/mol.K
T = temperature in Kelvin (K)
K = equilibrium constant for the reaction
Given;
K = 17.23
T = 371 K
R = 8.314 J/mol.K
Plugging in the given values into the equation;
ΔG°rxn = - (8.314 J/mol.K) × (371 K) * ln(17.23)
Using a calculator, we can calculate the natural logarithm of 17.23:
ln(17.23) ≈ 2.848
Plugging this value back into the equation;
ΔG°rxn = - (8.314 J/mol.K) × (371 K) × 2.848
ΔG°rxn ≈ - 6979 J/mol
The negative value of ΔG°rxn indicates that the reaction is spontaneous, as ΔG°rxn < 0.
Therefore, the reaction will occur spontaneously in the forward direction at 371 K when the equilibrium constant (K) is 17.23.
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at 300K, a sample of gas occupies 1.5 dm3.
Assuming the pressure remains constant, the volume of this gas at 600K would be?
What volume in milliliters of a 1.00 M solution of sodium hydroxide is required to
make 125 mL of a 0.0600 M solution?
7.50 mL
12.5 mL
16.7 mL
208 mL
S8 + 12O2 -------> 8SO3
If you start with 873.2 g of S8 and 859.3 g of O2, what mass of SO3 will be produced?
When you write the formula for sodium hydroxide, you do not have to put parentheses around the hydroxide polyatomic ion. However, when writing the formula for aluminum hydroxide, you must put parentheses around the hydroxide polyatomic ion. a) Write each formula. b) Explain why the parentheses are necessary for aluminum hydroxide.
(a). Sodium hydroxide: NaOH, aluminum hydroxide: [tex]Al(OH)_3[/tex]
(b). The parentheses are necessary for aluminum hydroxide because the hydroxide polyatomic ion has a subscript of 3, indicating that there are three hydroxide ions for every one aluminum ion.
a) The formula for sodium hydroxide is NaOH, and the formula for aluminum hydroxide is [tex]Al(OH)_3[/tex]
b) Aluminum hydroxide requires brackets because there are three hydroxide ions for every one aluminum ion, according to the hydroxide polyatomic ion's subscript of 3. Without the parentheses, it would be unclear whether the subscript of 3 applies to only the oxygen or to the entire hydroxide ion. By enclosing the entire hydroxide ion in parentheses and placing the subscript outside the parentheses.
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Determine the mass of 2.62 mol of iron(III) sulfate.
The molar mass of 2.62 mol of iron(III) sulfate is 1050.8 g.
Thus, the molar mass of iron(III) sulfate can be calculated by summing the atomic masses of its constituent atoms. The molar mass of a compound is the sum of the atomic masses of all the atoms present in a chemical formula of the compound which is then multiplied by the number of atoms of each element in the formula.
In iron(III) sulfate, the atomic mass of iron will be 111.70 g/mol. The atomic masses of Sulphur and oxygen will be 96.18 g/mol and 192.0 g/mol, respectively. Adding atomic masses of its constituent atoms will be 400.88 g/mol. Therefore, the molar mass of 2.62 mol of iron(III) sulfate is 1050.8 g.
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What is the amount (in mole) of sodium trioxocarbonate (IV) in 5.3g of the compound? (Na2CO3 = 106)
Answer:
To calculate the amount (in mole) of sodium trioxocarbonate (IV) in 5.3g of the compound , you can use the formula:
Amount (in moles) = Mass of substance / Molecular mass
The molecular mass of sodium trioxocarbonate (IV) (Na2CO3) is 106 g/mol, as given in the question. Substituting the values in the formula, we get:
Amount (in moles) = 5.3 g / 106 g/mol = 0.05 mol
Therefore, the amount (in mole) of sodium trioxocarbonate (IV) in 5.3g of the compound is 0.05 mol.
Note that the given molecular mass of Na2CO3 must be used to obtain the correct answe12 .
Explanation:
What is the density (in g/L) of CO2 in a 5.20 L tank at 760.0 torr and 39.0°C .
The tank's CO₂ density is 1.84 g/L.
How to calculate density?Use the ideal gas law to solve for the density of CO₂:
PV = nRT
where:
P = pressure = 760.0 torr
V = volume = 5.20 L
n = moles of CO2 (we don't know this yet)
R = gas constant = 0.08206 L·atm/K·mol
T = temperature = 39.0°C + 273.15 = 312.15 K
First, convert torr to atm:
760.0 torr ÷ 760 torr/atm = 1 atm
Rearrange the ideal gas law to solve for n:
n = PV/RT
n = (1 atm)(5.20 L)/(0.08206 L·atm/K·mol)(312.15 K)
n = 0.217 mol
Use the mass of CO₂ and the volume of the tank to find the density:
mass = n × molar mass
mass = 0.217 mol × 44.01 g/mol
mass = 9.57 g
density = mass/volume
density = 9.57 g/5.20 L
density = 1.84 g/L
Therefore, the density of CO₂ in the tank is 1.84 g/L.
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How many mL of a 5.00% (m/v) glucose solution will be needed to deliver 8.5 grams of glucose?
How much energy is released when 73 grams of water cools from 72 degrees Celsius to
30 degrees Celsius?
Determine the total pressure of a gas mixture that contains O
To determine the total pressure of a gas mixture that contains oxygen (O2), you need to know the partial pressure of the oxygen gas and the partial pressures of the other gases in the mixture. The total pressure of a gas mixture is equal to the sum of the partial pressures of all the gases in the mixture. This is known as Dalton’s Law of Partial Pressures. Can you provide more information about the gas mixture, such as the partial pressures or mole fractions of the gases in the mixture?
MARK ME BRAINLEISTA bag of frozen broccoli weighs 306.0 grams. You microwave it and notice a lot is steam so you weigh after microwaving and it is 275.0 grams. What happened to the percent mass of water? Show your work
There are different methods to calculate the concentration of a solution. Mass percentage is one among them. Mass percentage is mainly used to calculate the concentration of a binary solution. Here mass percent of water is 10.13.
Mass percentage of a particular component in a solution is equal to mass in grams of that component present per 100 g of the solution. For example, a 5% aqueous solution of urea means 5g of urea in 100 g of its aqueous solution.
Mass percentage = Mass of the component / Total mass of solution × 100
Mass of water = 306.0 - 275.0 = 31
% Mass = 31 / 306.0 × 100 = 10.13%
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a) If the chemical formula for iron (III) chloride is FeCl 3 , what is the chemical formula for iron (III) nitrate?b) If the chemical formula for lead (II) oxide is PbO, what is the chemical formula for lead (II) sulfate?c) If the chemical formula for silver chloride is AgCl, what is the chemical formula for silver nitrate?
a) The chemical formula for iron (III) nitrate is Fe(NO₃)₃. b) The chemical formula for lead (II) sulfate is PbSO₄. c) The chemical formula for silver nitrate is AgNO₃.
In chemical nomenclature, the Roman numeral in the name of the compound indicates the oxidation state of the metal ion. To determine the chemical formula of a compound, it is important to balance the charge of the ions in the compound. In the case of iron (III) nitrate, the iron ion has a +3 charge and the nitrate ion has a -1 charge, so it takes three nitrate ions to balance the charge of the iron ion.
In the case of lead (II) sulfate, the lead ion has a +2 charge and the sulfate ion has a -2 charge, so it takes one lead ion and one sulfate ion to balance the charges. Similarly, in the case of silver nitrate, the silver ion has a +1 charge and the nitrate ion has a -1 charge, so it takes one of each to balance the charges.
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A student used a video camera to record another student dropping a marble through water in a graduated cylinder. The students watched the video in slow motion and made the observations shown below. During which part or parts of the marble’s fall did the marble experience unbalanced forces?
Parts B and C of the marble's fall did the marble experience unbalanced forces. Option 4 is correct.
A force is a push or pull (interaction) which changes the momentum of an object, either stationary or in motion when unopposed. All objects experience different forces depending on their environment. When immersed in fluids, unbalanced forces of one upward moving force tends to cancel the gravity force moving downward on a sinking object causing deceleration to a constant sinking speed.
This upward moving force is called as Buoyant force. This is where at part A, the object will experiences a balanced force of gravity which accelerates due to the absence of an opposing force acting upwards on the object. At part B, the speed of the sinking object decreases due to an unbalanced force that cancels the acceleration by the buoyant force. Once the sinking object’s acceleration is cancelled, its sinking speed turns constant at part C.
Hence, 4. is the correct option.
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--The given question is incomplete, the complete question is
"A student used a video camera to record another student dropping a marble through water in a graduated cylinder. The students watched the video in slow motion and made the observations shown below. During which part or parts of the marble's fall did the marble experience unbalanced forces? (1) Part A only (2) Parts A and B only (3) Part C only (4) Parts B and C only."--
