What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl

Answers

Answer 1

Answer:

Solution given:

1 mole of KCl[tex]\rightarrow [/tex]22.4l

1 mole of KCl[tex]\rightarrow [/tex]74.55g

we have

0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g

74.55g of KCl[tex]\rightarrow [/tex]22.4l

10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

Answer 2

[tex]\:[/tex]

1 mole of KCl → 22.4l

1 mole of KCl → 74.55g

we have

0.14 mole of KCl → 74.55*0.14=10.347g

74.55g of KCl  → 22.4l

10.347 g of KCl → 22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.


Related Questions

Rank the compounds NH3, CH4, and PH3 in order of decreasing boiling point. Choices: A) NH3 > CH4 > PH3 B) CH4 > NH3 > PH3 C) NH3 > PH3 > CH4 D) CH4 > PH3 > NH3 E) PH3 > NH3 > CH4

Answers

Answer:

C) NH3 > PH3 > CH4

Explanation:

The boiling point of a substance depends on the nature of intermolecular interaction between the molecules of the substance. The greater the magnitude of intermolecular interaction between the molecules of the substance, the higher the boiling point of the substance.

Both NH3 and PH3 have intermolecular hydrogen bonding between their molecules. However, since nitrogen is more electronegative than phosphorus, the magnitude of intermolecular hydrogen bonding in NH3 is greater than in PH3 hence NH3 has a higher boiling point than PH3.

CH4 molecules only have weak dispersion forces between them hence they exhibit the lowest boiling point.

Chlorine radicals perform the first propagation step:

a. in comparison to bromine radicals.
b. radicals form easily in the presence of chlorine radicals.
c. Subsequently, the resulting radicals can react with bromine in a second propagation step to yield monobrominated products.

Answers

Answer:

b. radicals form easily in the presence of chlorine radicals.

Explanation:

Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."

This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.

Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."

All of the following statements concerning crystal field theory are true EXCEPT Group of answer choices in an isolated atom or ion, the five d orbitals have identical energy. low-spin complexes contain the maximum number of unpaired electrons. in low-spin complexes, electrons are concentrated in the dxy, dyz, and dxz orbitals. the energy difference between d orbitals often corresponds to an energy of visible light. the crystal field splitting is larger in low-spin complexes than high-spin complexes.

Answers

Answer:

low-spin complexes contain the maximum number of unpaired electrons.

Explanation:

In the crystal field theory, the magnitude of crystal field splitting and the pairing energy determines whether a complex will be low spin or high spin.

Low spin complexes often have greater magnitude of crystal field splitting energy than low spin complexes.

High spin complexes have maximum number of unpaired electrons(most of the electrons are unpaired) while low spin complexes have a minimum number of unpaired electrons in a complex(most of the electrons are paired).

Compare the total number of modes for 4 moles of a monoatomic gas and 1 mole of a gas consisting of linear triatomic molecules (such as CO2 gas). If these two gases, initially at difference temperatures, were placed in the same container and allowed to reach equilibrium, which gas (if any) would have a greater change in temperature

Answers

Answer:

23

Explanation:

Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Given that the atomic radii of H and Br are 37.0 pm and 115 pm , respectively, predict the upper limit of the bond length of the HBr molecule. Express your answer to three significant figures and include the appropriate units. View Available Hint(s)for Part C

Answers

Answer:

The answer is "152 pm".

Explanation:

The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.

The atomic radius of [tex]H= 37.0 \ pm[/tex]

The atomic radius of [tex]Br = 115.0 \ pm[/tex]

[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]

                    [tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32 . Calculate the Ka for the acid.

Answers

Answer:

1.94 × 10⁻³

Explanation:

Step 1: Calculate the concentration of H⁺ ions

We will use the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M

Step 2: Calculate the acid dissociation constant (Ka) of the acid

For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.

Ka = [H⁺]²/Ca

Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³

Identify the compound that possesses a permanent dipole. Please choose the correct answer from the following choices, and then select the submit answer button. Answer choices acetone, (CH3)2CO cyclohexane, C6H12 pentane, C5H12 methane, CH4.

Answers

Answer:

acetone, (CH3)2CO cyclohexane are the compound that possesses a permanent dipole

Explanation:

Permanent dipole describes the partial charge separation that can occur within a molecule along with the bond dat form between 2 different atoms

How are radioactive isotopes used to diagnose thyroid problems?
Answers

A.
Production of radioactive iodine by the thyroid is measured.

B.
Uptake of radioactive iodine by the thyroid gland is measured.

C.
Destruction of damaged cells by radioisotopes is measured.

D.
Absorption of gamma rays from ingested radioisotopes is measured.
Question

Answers

Answer:

A

Explanation:

uptakw of raduoactive iofine by the thyroid gland js measurd

Sodium acetate is produced by the reaction of baking soda and vinegar. The resultant solution is then heated until it becomes saturated and allowed to cool. As a result, the solution has become supercooled. Upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes. Its molar enthalpy of fusion is 35.9 kJ/mol. How much thermal energy would be released by 276.0 g of sodium acetate trihydrate (molar mass

Answers

Answer: The thermal energy that would be released by 276.0g of sodium acetate trihydrate is 71.8kJ.

Explanation:

Supercooling is the process of lowering the temperature a liquid below its freezing point, without it becoming solid. A liquid below its freezing point will crystallize in the presence of a seed crystal because it serves as a structure for formation of crystals. From the question,

The given mass of sodium acetate trihydrate

(CH3COONa.3H2O)= 276.0g

Molar mass of sodium acetate

trihydrate= 136.08g/mol

Thermal heat of fusion of sodium acetate

trihydrate = 35.9 kJ/mol

From the given mass the number of moles present= 276.0/ 136.08

= 2.0moles

Therefore the heat (thermal) energy of the given mass of sodium acetate

trihydrate = 2.0 × 35.9

= 71.8kJ

Therefore, upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes.

A 13.4 mL sample of CO2 gas was collected in an experiment.
What is this volume in liters (L)? Use significant figures, do NOT include the units.

Answers

Explanation:

here's the answer to your question

What trends were seen in medeleevs periodic table

Answers

Answer:

groups are based on how many electrons to become stable

Explanation:

What is the oxidation state of rubidium (Rb)?

A. +1
B. -2
C. +2
D. -1​

Answers

Answer:

The answer is A. +1

Explanation:

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is

Answers

Answer:

p'PCl3 =  6.8 torr

p'Cl2 =26.4 torr

p'PCl5 =223.4 torr

Explanation:

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is

PCl3(g) + Cl2(g) ---> PCl5(g)

Calculate the new partial pressures after equilibrium is reestablished. [in torr]

pPCl3

pCl2

pPCl5

Step 1: Data given

Partial pressure before adding chlorine gas:

Partial pressure of PCl5 = 217.0 torr

Partial pressureof PCl3 = 13.2 torr

Partial pressureof Cl2 = 13.2 torr

A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing

Step 2: The equation

PCl3(g)+Cl2(g) ⇔ PCl5(g)

Step 3: The expression of an equilibrium constant before adding chlorine gas

Kp = pPCl5 / (pPCl3 * pCl2)

Kp = 217.0 / (13.2 * 13.2)

Kp =  1.245

Step 4:  The expression of an equilibrium constant after adding chlorine gas

Partial pressure of PCl5 = 217.0 torr

Partial pressure of PCl3 = 13.2

Partial pressure of Cl2 = TO BE DETERMINED

Step 5: The total pressure of the system

Ptotal = pPCl5 + pPCl3 + pCl2

263.0 torr = 217.0 torr + 13.2 torr + pCl2

pCl2 = 263.0 - 217.0 -13.2 = 32.8 torr

Step 6: The initial pressure

The equation: PCl3(g)+Cl2(g) ⇔ PCl5(g)

pPCl3 = 13.2 torr

pCl2 = 32.8 torr

pPCl5 = 217.0 torr

Step 7: The pressure at the equilibrium

p'PCl3 = (13.2 -x) torr

p'Cl2 = (32.8 - x) torr

p'PCl5 = (217.0 + x) torr

Step 8: The equilibrium constant

'Kp =  p'PCl5 / (p'PCl3 * p'Cl2)

1.245 = (217.0+x) / ((13.2-x)(32.8-x)

x = 6.40 torr

p'PCl3 = 13.2 -6.40 = 6.8 torr

p'Cl2 = 32.8 - 6.40 =26.4 torr

p'PCl5 = 217.0 + x) 6.4 = 223.4 torr

A certain watch’s luminous glow is due to zinc sulfide paint that is energized by beta particles given off by tritium, the radioactive hydrogen isotope 3 H, which has a half-life of 12.3 years. This glow has about 1/10 of its initial brightness. How many years old is the watch? g

Answers

Answer:

The watch is 40.9 years old.

Explanation:

To know how many years old is the watch we need to use the following equation:

[tex] I_{(t)} = I_{0}e^{-\lambda t} [/tex]   (1)

Where:

[tex]I_{(t)}[/tex]: is the brightness in a time t = (1/10)I₀

[tex]I_{0}[/tex]: is the initial brightness

λ: is the decay constant of tritium

The decay constant is given by:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]   (2)

Where:

[tex]t_{1/2}[/tex]: is the half-life of tritium = 12.3 years

By entering equation (2) into (1)  we have:

[tex] I_{(t)} = I_{0}e^{-\lambda t} = I_{0}e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]

[tex] \frac{I_{(t)}}{I_{0}} = e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]

By solving the above equation for "t" we have:

[tex] ln(\frac{I_{(t)}}{I_{0}}) = -\frac{ln(2)}{t_{1/2}}t [/tex]

[tex] t = -\frac{ln(\frac{I_{(t)}}{I_{0}})}{\frac{ln(2)}{t_{1/2}}} = -\frac{ln(\frac{1}{10})}{\frac{ln(2)}{12.3}} = 40.9 y [/tex]

Therefore, the watch is 40.9 years old.

 

I hope it helps you!

Predict whether reactants or products will be favored at equilibrium for the below reaction.

Kp= 2.2 x 10^6 at 298K
2COF2 (g) + ⇌ CO2(g) + CF4(g)

Answers

Answer:

The products will be favored at equilibrium.

Explanation:

The balanced chemical equation for the reaction is the following:

2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)

The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).

The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:

[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]

Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²

So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).

Therefore, products will be favored at equilibrium at 298 K.

Dylan has a coworker who is always showing up late and then not finishing his work on time . It's frustrating the other members of the team . What can he do that might help the situation ? a ) Complain about the coworker to other team members b ) Ask his coworker if he understands his job responsibilities c ) Tell his boss that the coworker is slacking off d ) Complete his coworker's work for him

Answers

The answer should be C I think that’s the best answer but I’m not sure.

For each reaction, write the chemical formulae of the oxidized reactants.

a. ZnCl2 (aq) + 2Na(s) → Zn(s) + 2NaCl(aq)
b. Al(s) + FeBrz (aq) → AlBrz (aq) + Fe(s)
c. FeSO4 (aq) + Zn (s) → Fe(s) + ZnSO4(aq)

Answers

Answer:

a. Na(s); b. Al(s); c. Zn(s)

Explanation:

Let's consider the following redox reactions.

a. ZnCl₂ (aq) + 2 Na(s) → Zn(s) + 2 NaCl(aq)

Na is oxidized because its oxidation number increases from 0 to +1 (in NaCl) whereas Zn is reduced because its oxidation number decreases from 2+ (in ZnCl₂) to 0.

b. Al(s) + FeBr₃ (aq) → AlBr₃ (aq) + Fe(s)

Al is oxidized because its oxidation number increases from 0 to +3 (in AlBr₃) whereas Fe is reduced because its oxidation number decreases from 3+ (in FeBr₃) to 0.

c. FeSO₄ (aq) + Zn(s) → Fe(s) + ZnSO₄(aq)

Zn is oxidized because its oxidation number increases from 0 to +2 (in ZnSO₄) whereas Fe is reduced because its oxidation number decreases from 2+ (in FeSO₄) to 0.

how many moles of CO2 are formed from 3.0 mol of C2H2

Answers

Answer:

50.0 moles C02

Explanation:

First write down the CORRECTLY balanced equation. NOTE: The equation you provide is incorrect.

2C2H2(g) + 5O2(g) ==> 4CO2(g) + 2H2O(g)  CORRECT EQUATION

Next, look at the stoichiometric ratio of C2H2 to CO2.  You can see it is 2 moles C2H2 produces 4 moles CO2.

Thus, 25.0 moles C2H2 x 4 moles CO2/2 moles C2H4 = 50.0 moles CO2

How many grams of NaCl (MM = 58.44g/mol) are in 250mL of a 0.75 molar solution?

Answers

Answer:

[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]

Explanation:

We are asked to find how many grams of sodium chloride are in a solution.

1. Moles of Solute

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]

We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.

There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.

[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]

Now we know the molarity and the liters of solution, but the moles of solute are unknown.

molarity = 0.75 mol NaCl/L moles of solute =x liters of solution = 0.25 L

Substitute the values into the formula.

[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]

We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]

The units of liters cancel.

[tex]0.250 * 0.75 \ mol \ NaCl[/tex]

[tex]\bold {0.1875 \ mol \ NaCl}[/tex]

2. Grams of Solute

Now that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.

Set up a ratio so we can convert using dimensional analysis.

[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

Multiply by the number of moles we calculated.

[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

The units of moles of sodium chloride cancel.

[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]

[tex]\bold {10.9575 \ g \ NaCl}[/tex]

3. Round using Significant Figures

The original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.

[tex]11 \ g \ NaCl[/tex]

There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.

Radon-220 undergoes alpha decay with a half-life of 55.6 s.?
Assume there are 16,000 atoms present initially and calculate how many atoms will be present at 0 s, 55.6 s, 111.2 s, 166.8 s, 222.4 s, and 278.0 s (all multiples of the half-life). Express your answers as integers separated by commas.
Calculate how many atoms are present at 50 s, 100 s, and 200 s (not multiples of the half-life).

Answers

The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.

The equation below gives the number of atoms present at time t

[tex]N=Noe^-kt[/tex]

N = Number of atoms present at time t

No = Number of atoms initially present

k = decay constant

t = time taken

Given that;

t1/2 = 0.693/k

where t1/2 = half life

k = 0.693/t1/2

k = 0.693/ 55.6 s

k = 0.0125 s-1

Substituting values;

N = 16,000 e^-0.0125(0)

N = 16,000 atoms

At 50 s

N = 16,000 e^-0.0125(50)

= 8564 atoms

At 100 s

N = 16,000 e^-0.0125(100)

= 4584 atoms

At 200 s

N = 16,000 e^-0.0125(200)

= 1313 atoms

https://brainly.com/question/2998270

The enthalpy of formation for CO2 (s) and CO2 (g) is: -427.4 KJ/mole and -393.5 KJ/mole, respectively. The sublimation of dry ice is described by CO2 (s) → CO2 (g).

The enthalpy needed to sublime 986 grams of CO2 is:
(a) 181.5 Kcal
(b) 611.7 Kcal
(c) 248.3 Kcal
(d) 146.2 Kcal

Answers

Answer:

a. 181.5 kcal

Explanation:

Step 1: Calculate the enthalpy of the process (ΔH°).

Let's consider the following process.

CO₂(s) → CO₂(g)

We can calculate the enthalpy of the process using the following expression.

ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)

ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))

ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ

According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.

Step 2: Convert 986 g of CO₂ to moles

The molar mass of CO₂ is 44.01 g/mol.

986 g × 1 mol/44.01 g = 22.4 mol

Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂

22.4 mol × 33.9 kJ/1 mol = 759 kJ

We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.

759 kJ × 1 kcal/4.184 kJ = 181.5 kcal

what is the charge on the Mn ions in Mn2o3? 1+, 2+, 3+,3-,4+?

Answers

hey here’s your answer hope this helps you!!!!

importance of hematology​

Answers

Answer:

Haematology is the specialty important for the diagnosis and management of a wide range of benign and malignant disorders of the red and white blood cells, platelets and the coagulation system in adults and children.

Write a balanced half-reaction for the oxidation of manganese ion (mn2 ) to solid manganese dioxide (mno2) in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

Explanation:

Step 1: Data given

The oxidation number of manganese ion (Mn2+ ) is +2

The oxidation number of manganese dioxide  is +(MnO2)4

This means the oxidation number from Mn will go from +2 to +4, since it's increased, this is an oxidation reaction

Mn2+(aq)  ⇒ MnO2(s)

We have to balance both sides. Mn is already the same. But on the right side we have O atoms. T obalance both sides we have to add O atoms to the left side. This by adding 2x H2O

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s)

Now the amount of O atoms is balanced, but we have H- atoms at the left side. To balance we have to add 4 H atoms to the right side

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq)

Now the amount of atoms is balanced at both sides. We also have to check if the charge on both sides is the same.

Since the left side has a charge of +2, and right has a charge of +4, we have to add 2 electrons to balance this.

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

Balance the following reaction:

_______ CO₂ + _______ H₂O + heat ↔ _______ C₆H₁₂O₆ + _______ O₂

Please explain!
*Note: If any of the coefficients are the number one. Please, write "1" in the space. Thanks!

Answers

Answer:

6CO2+6H2O+heat" C6H12O6+6O2

the best way to balance a chemical reaction is to start with balancing the hydrogen followed by the other elements then lastly oxygen.so in this case if you put a 6 in front of carbon dioxide,water and oxygen you will definitely balance it.cause at the first side you have 6 carbons similar to the product,12 oxygen similar to the product and 18 oxygen similar to the products.

I hope this helps

Answer:

Explanation:

I saw this after answering your other question on the same reaction.

To balance the chemical reaction, look at the reactants and products. As O is part of both products, focus on C and H instead.

On the products side, 1 C6H12O6 has 6 C and 12 H. So that requires the same numbers of C and H on the reactant side because of mass conservation.

That gives 6 CO2 and 6 H2O as the reactants. Counting the number of O in the reactants, there are 6*2 + 6 = 18 O. Subtracting the 6 O in C6H12O6, that leaves 12 O so there are 12/2 = 6 O2 in the products.

Combining the numbers above, the balanced equation is:

___6___ CO₂ + ___6___ H₂O + heat ↔ ___1___ C₆H₁₂O₆ + ___6___ O₂

there is 3.5 g of fat in a granola bar. You determine the fat content to be 4.0 G in the lab. What is the percent error

Answers

Answer:

[tex]error = 4.0 - 3.5 = 0.5 \\ \\ percent \: error = \frac{0.5}{3.5} \times 100 \\ \\ = 14.29\% [/tex]

Hydrogen chloride decomposes to form hydrogen and chlorine, like this:

2HCl(g) + H2(g) â Cl2(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, hydrogen, and chlorine has the following composition:

compound pressure at equilibrium
HCl 84.4 atm
H2 77.9 atm
Cl2 54.4

Required:
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answers

Solution :

Given :

Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm

Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm

Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm

Reaction :

[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]

Using equilibrium concept,

[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]

[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]

[tex]$k_p=0.594$[/tex]

[tex]k_p=0.59[/tex]  (in 2 significant figures)

or [tex]k_p=5.9 \times 10^{-1}[/tex]

Write the balanced equation for the hydration of CuSO4CuSO4. Indicate the physical states using the abbreviations (ss), (ll), or (gg) for solid, liquid, or gas, respectively. Use (aqaq) to indicate the aqueous phase. Indicate appropriate charges on negative and positive ions if they are formed.

Answers

Answer:

CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)

Explanation:

Hydration is the process by which anhydrous CuSO4 acquires molecules of water of crystalization to form the pentahydrate.

The water of crystalization becomes attached go the crystals of the CuSO4 to form the hydrated salt.

Beginning with solid anhydrous CuSO4 we have;

CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)

The graph below shows how the temperature and volume of a gas vary when
the number of moles and the pressure of the gas are held constant. What
happens to the
temperature of a gas as its volume increases?
A. The temperature decreases
B. The temperature increases
C. The temperature remains the same.
D. The temperature doubles.

Answers

Answer:

C

Explanation:

because it remains the same

Classify these bonds as ionic, polar covalent, or nonpolar covalent. You are currently in a sorting module.
Ionic Polar Covalent Nonpolar covalent
C-O , Mg-F , Cl-Cl

Answers

Answer:

C-O: polar covalent

Mg-F: ionic

Cl-Cl: nonpolar covalent

Explanation:

Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.

Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.

In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.

In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.

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