Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE
Answer:
The answer is o2
Explanation:
I took the test
PLEASE HELP!!
You are performing an experiment that involves the electrolysis of gold (I) bromide, also know as AuBr. You know that gold is less reactive than hydrogen. Which of the following would be the product of the reaction?
A. Hydrogen gas
B. Gold bromide
C. Oxygen gas
D. Pure gold
Answer:
D. Pure gold
Explanation:
Hello,
In this case, since gold, as a heavy metal, is said to be less reactive than hydrogen, when it undergoes electrolysis process when forming a salt, due to the action of the electric current, we can appreciate the formation of a layer of gold on the surface of the cathode via a reduction half-reaction from gold (I) to metallic gold:
[tex]Au^++1e^-\rightarrow Au^0[/tex]
Thereby, D. Pure gold is formed as the product of the reaction.
In contrast, more reactive metals than hydrogen such as sodium or potassium, will remain in solution so the hydrogen converted to hydrogen gas.
Best regards-
What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic
weight of Ca is 40.1 and the atomic weight of Br is 79.9.
A) 452.3 g
B) 53.04 g
C) 44.2 g
D) 88.4 g
Answer:
Below
Explanation:
Let n be the quantity of matter in the Calcium Bromide
● n = m/ M
M is the atomic weight and m is the mass
M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)
M = 40.1 + 2×79.9
● 0.422 = m/ (40.1+2×79.9)
●0.422 = m/ 199.9
● m = 0.422 × 199.9
● m = 84.35 g wich is 88.4 g approximatively
88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.
What do you mean by mass ?Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .
To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,
Let n be the quantity of matter in the Calcium Bromide
M is the atomic weight and m is the mass
n = m/ MM of CaBr2 is the sum of the atomic weight of its components
Mass of Ca = 40.1 , Mass of Br = 79.9
M = 40.1 + 2×79.9
0.422 = m/ (40.1+2×79.9)
0.422 = m/ 199.9
m = 0.422 × 199.9
m = 84.35 g which is 88.4 g approximatively .
Thus ,88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .
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A particular reaction at constant pressure is spontaneous at 390K. The enthalpy change for this reaction is +23.7kJ. What can you conclude about the sign and magnitude of ΔS for this reaction?a. smallb. largec. + smalld. + largee. 0.0
Answer:
+ small
Explanation:
The entropy is obtained from;
∆S= ∆H/T
Where;
∆S= entropy of the system
∆H= enthalpy if the system = +23.7 KJ
T= absolute temperature of the system = 390 K
∆S= 23.7 ×10^3/390 = 60.8 JK^-
There is a small positive change in entropy.
If 75.4 J of energy is absorbed by 0.25 mol of CCl4 at constant pressure, what is the change in temperature? The specific heat of CCl4 is 0.861 J/g·°C.
Answer:
ΔT = 2.28°C
Explanation:
Heat, H = 75.4J
Number of moles = 0.25 mol
Specific heat capacity, c = 0.861 J/g·°C
Change in temperature, ΔT = ?
These quantities are related by the following equation;
H = mc ΔT
Mass, m = Number of moles * Molar mass
m = 0.25mol * 153.82 g/mol
m = 38.455g
S back to the equation;
H = mc ΔT
Substituting the values;
75.4 = 38.455 * 0.861 * ΔT
ΔT = 75.4 / 33.11
ΔT = 2.28°C
The change in temperature is 2.28 °C
First, we will determine the mass of CCl₄ absorbed
From the given information,
Number of moles of CCl₄ absorbed = 0.25 mol
Using the formula
Mass = Number of moles × Molar mass
Molar mass of CCl₄ = 153.82 g/mol
∴ Mass of CCl₄ absorbed = 0.25 × 153.82
Mass of CCl₄ absorbed = 38.455 g
Now, using the formula
Q = mcΔT
Where Q is the quantity of heat
m is the mass
c is the specific heat of substance
and ΔT is the change in temperature
From the given information
Q = 75.4 J
c = 0.861 J/g.°C
Putting the parameters into the formula, we get
75.4 = 38.455 × 0.861 ×ΔT
75.4 = 33.109755 × ΔT
∴ ΔT = 75.4 ÷ 33.109755
ΔT = 2.28 °C
Hence, the change in temperature is 2.28 °C
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which of the following is an example of precipitate a. liquid evaporating into gas b. a solid form from a frozen liquid c. a chunky solid form from 2 liquids combining d. a liquid solution that contains 2 substances
Answer:
B and C
Explanation:
The google definition of precipitate is "a solid formed by a change in a solution, often due to a chemical reaction or change in temperature that decreases solubility of a solid". So in this case, the "solid formed from a frozen liquid" and "a chunky solid form from 2 liquids combining" are both examples of a precipitate.
The correct answer is C, a chunky solid form from 2 liquids combining.
In the laboratory, we usually mix two chemicals. When we mix the chemicals, we expect them to interact in one way or another. The interaction of those chemicals is known as a chemical reaction.
When we mix two liquid chemicals and they interact with each other to yield a solid product, we say that a chemical reaction has occurred and that such chemical reaction has produced a precipitate.
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A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculate the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol. He next tried the same experiment with 100 ml of 1.0 M HCl and 100 ml of 1.0 M NaOH. The calculated molar enthalpy change of reaction for his second trial was:
Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.
Explanation:-
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]
Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]
Given for second trial:
[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]
[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]
0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat = 54 kJ
0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release heat =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]
Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.
Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
For which one of the following reactions will the enthalpy change be approximately equal to the internal energy change?
A. H2 + I2 → 2HI
B. PCl5(g) → PCl3(g) + Cl2
C. 2H2O2 → 2H2O2 + O2
D. C(s) + O2(g) → CO2(g)
Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]
Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:
[tex]\Delta H=\Delta E+\Delta(PV)[/tex]
So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.
Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.
Analysing each reaction above:
Reaction A has the same number of moles in reagents and products, so enthalpy change and internal energy change will be equal;
Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.
Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.
When balancing redox reactions under basic conditions in aqueous solution, the first step is to:________.
a. balance oxygen
b. balance hydrogen
c. balance the reaction as though under acidic conditions
d. none of the above
Answer:
When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Explanation:
Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.
The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.
The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:
In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.
In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-
So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Answer:
c. balance the reaction as though under acidic conditions
Explanation:
When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.
If an individual proton has mass 1.007825 amu, and an individual neutron has mass 1.008665 amu, what's the calculated mass of a neptunium-236 nucleus? options: A) 237.92482 amu B) 236.99873 amu C) 237.96682 amu D) 237.04817 amu
Answer:
C) 237.96682 amu
Explanation:
The symbol for neptunium-236 is given as;
²³⁶₉₃Np
This element has 93 protons and (236 - 93 = 143) neutrons.
Mass Number =Total mass of Protons + Total mass of neutrons
Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu
Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu
Mass = 144.239095 + 93.727725 = 237.96682 amu
Correct option is option C.
For the following set of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain constant.
V=2.91 L at 23.0 °C
V= 4.20 L at ? °C
Answer:
155 °C
Explanation:
Step 1: Given data
Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?Step 2: Convert the initial temperature to Kelvin
We will use the following expression.
K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K
Step 3: Calculate the final temperature
Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 4.20 L × 296.2 K/2.91 L
T₂ = 428 K
Step 4: Convert the final temperature to Celsius
We will use the following expression.
°C = K - 273.15 = 428 - 273.15 = 155 °C
What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?
Answer:
The rate would have doubled
Explanation:
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
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When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many
The given question is incomplete.
The complete question is:
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?
Answer: 4 grams of methane were needed for the reaction
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]
Given: mass of oxygen = 16 g
Mass of carbon dioxide = 11 g
Mass of water = 9 g
Mass of products = Mass of carbon dioxide + mass of water = 11 g +9 g = 20 g
Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g
As mass of reactants = mass of products
mass of methane + 16 g= 20 g
mass of methane = 4 g
Thus 4 grams of methane were needed for the reaction
Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
Which option draws the correct conclusion from the following case study?
A patient with sickle-cell anemia and a fever goes to the emergency room and is given Tylenol to reduce
the fever. The patient has seizures and dies after taking the Tylenol. The physician writes up this case as
an interesting outcome for a patient with sickle-cell anemia.
The case study's validity is obvious because it describes a real-life situation.
The case study was influenced by bias, and led to incorrect conclusions being drawn
The case study was not intended to produce a generalized conclusion about treatment
Upon reading this case study, physicians should stop treating sickle cell patients with fevers using Tylenol
Answer:
I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong
Explanation:
Answer: options B
Explanation:
The intermolecular forces present in CH 3NH 2 include which of the following? I. dipole-dipole II. ion-dipole III. dispersion IV. hydrogen bonding
Answer:
I. dipole-dipole
III. dispersion
IV. hydrogen bonding
Explanation:
Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.
London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.
Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.
Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.
Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.
Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.
Answer:
The intermolecular forces present in CH_3NH_2 includes
II. (ion-dipole) and IV. (hydrogen bonding)Explanation:
The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)
It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.
Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.
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When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
What are the products in the following chemical reaction Pb(NO3)+KCI
Answer:
The products are KNO3 + PbCl2.....
Espero que te sirva.
is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous compound is left. Further analysis shows that the percentage composition of the anhydrate is 21.90% Ca, 43.14% Se, and 34.97% O.. (Hint: Treat the anhydrous compound and water just as you have treated elements in calculating in the formula of the hydrate.) (Use an asterisk to enter the dot in the formula. If a subscript is 1, omit it.) Find the empirical formula of the anhydrous compound. Find the empirical formula of the hydrate.
Answer:
The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.
Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams
The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,
Moles of water = weight of water/molecular weight
= 42.1 grams / 18 = 2.3
The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,
= 214.2 * 0.2190 = 46.91 grams
The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,
= 214.2 * 0.4314 = 92.40 grams
The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,
= 214.2 * 0.3497 = 74.91 grams
The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17
The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,
92.40/78.96 = 1.17
The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,
74.91/15.999 = 4.68.
Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O
If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
How many grams is 5.8 moles of hydrochloric acid (HCI)?
Answer to the nearest 0.01 g.
Answer:
211.47 grams
Explanation:
We need to set up a dimensional analysis to solve this problem by converting from moles to grams.
First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:
1.01 + 35.45 = 36.46 g/mol
We have 5.8 moles of HCl, so multiply by its molar mass:
(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g
The answer is thus 211.47 grams.
~ an aesthetics over
Answer:
[tex]\large\boxed{211.47}\\[/tex] grams
Explanation:
First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.
Hydrogen has an atomic weight of 1.00784 and Chlorine has an atomic mass of 35.453.Add those two values together - 1.00784 + 35.453 = 36.46084Multiply this value by 5.8 (one mole is equivalent to the atomic mass of the compound) - 5.8 x 36.46084 = 211.472872Round to the nearest 0.01 gram - 211.47[tex]\large\boxed{211.47}[/tex] is the final answer.
What is the initial temperature (°C) of a system that has the pressure decreased by 10 times while the volume increased by 5 times with a final temperature of -123°C?
Answer:
27°C or 300K
Explanation
We were told that the pressureof the system decreased by 10 times implies that P2= P1/10
Where P2=final pressure
P1= initial pressure
Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1
Where T2= final temperature =-123C= 273+(-123C)=150K
T1= initial temperature
But from gas law
PV=nRT
As n and R are constant
P1V1/T1 = P2V2/T2
T1= P1V1T2/P2V2
T1=2×T2
T1=2×150
T1=300K
=300-273
=27°C
the initial temperature (°C) of a system is 27°C
i) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the specific structural information you get from each method.
IR
UV-VIS
NMR
Mass Spec
delete please .....................................
A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
Which of the following ionic lattices would have the highest melting point?
A. Potassium oxide
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B. Boron nitride
C. Beryllium oxide
D. Lithium chloride
Answer:
I think, berryllium oxide, is answer.Explanation:
Hope it helps you....The ionic lattices would have the highest melting point Potassium oxide. option A is correct.
what is ionic lattice?An ionic compound is a giant structure of ions. The ions have a regular, repeating arrangement called an ionic lattice. The lattice is formed because the ions attract each other and form a regular pattern with oppositely charged ions next to each other.
Ionic compounds are held together by electrostatic forces between oppositely charged ions.
These forces are usually referred to as the ionic lattice contains such a large number of ions, that a lot of energy is needed to overcome this ionic bonding so ionic compounds have high melting and boiling points.
therefore, sodium oxide has the highest melting point. option A is correct
Learn more about ionic lattice here:
https://brainly.com/question/27549049
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What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?
Answer:
[tex]V_2=1.17L[/tex]
Explanation:
Hello,
In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:
[tex]V_1C_1=V_2C_2[/tex]
Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:
[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]
Best regards.
Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets: How many moles of each of the gaseous products are produced when 20.1 g of pure hydrazine is ignited in the presence of 20.1 g of pure oxygen
Answer:
[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]
Explanation:
Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:
[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]
Now we can balance the reaction:
[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]
In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):
[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]
[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]
In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].
With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]
We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]
I hope it helps!
Draw a picture of what you imagine solid sodium chloride looks like at the atomic level. (Do NOT draw Lewis structures.) Make sure to include a key. Then describe what you've drawn and any assumptions you are making.
Answer:
Kindly check the explanation section.
Explanation:
PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.
The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and
Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.
In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.
The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.
In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.
The expected structure of the NaCl would be the image attached below.
The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.
For more information about the structure of NaCl, refer to the link:
https://brainly.com/question/2729718
