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olchik [2.2K]
3 years ago
10

*30 POINTS* ANSWER THESE 5 MULTPLE CHOICE QUESTIONS!!! I'LL GIVE 30 POINTS AND A BRAINIEST:)

Chemistry
2 answers:
IgorLugansk [536]3 years ago
6 0

C, A, A, B, D. Hope this is correct! Have an awesome day!

erastova [34]3 years ago
3 0

Answer:

Do you need help still.........

Explanation:

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38 points!! >:D
Sav [38]

Answer: which*

Explanation:

8 0
3 years ago
Read 2 more answers
How is Boyle’s law different from Charles’s law?
katrin [286]

Answer:

In Charles law, temperature and volume of the gas are kept at constant pressure. Where as in Boyle's law, pressure and volume of the gas are kept at a constant temperature. In Boyle's law, pressure and volume vary inversely where as, in Charles law, pressure and volume vary directly.

3 0
3 years ago
Please help I need the answer by today and it is very important! Which of the following statements best describes why the mass o
Juliette [100K]

D, because matter does not change in space but unlike gravity, which changes, matter, is the mass of it, which will stay the same unless there is a chemical change or, physical change, but not in this case, so the answer you are looking for is D. hope this enlarged your understanding and knowledge.

6 0
3 years ago
How many grams of xenon are required to produce 10.0 g of XeF6?
Vaselesa [24]
Moles XeF6 = 10.0g/ 245.28 g/ mol=0.0408
The ratio between F2 and XeF6 is 3:1
Moles F2 required = 3 x 0.0408=0.122
Mass F2 = 0.122 mol x 37.9968 g/ mol=4.64g
6 0
3 years ago
[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)
pav-90 [236]

<u>Answer:</u>

<u>For A:</u> The expression for K_{eq} is given below.

<u>For B:</u> The value of K_{eq} at 25°C is 0.0185512

<u>For C:</u> The value of K_{eq} at 65°C is 0.2887886

<u>For D:</u> The reaction is endothermic in nature.

<u>Explanation:</u>

  • <u>For A:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

For the given chemical reaction:

[Co(H_2O)_6]^{2+}(aq.)+4Cl^-(aq.)\rightleftharpoons [CoCl_4]^{2-}(aq.)+6H_2O(l)

The expression of K_{eq} for above equation without the concentration of liquid water is:

K_{eq}=\frac{[CoCl_4]^{2-}}{[Co(H_2O)_6]^{2+}[Cl^-]^4}      ......(1)

The expression is written above.

  • <u>For B:</u>

We are given:

[CoCl_4]^{2-}=0.0334612M

[Co(H_2O)_6]^{2+}=0.966539M

[Cl^-]=1.86616M

Putting values in equation 1, we get:

K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512

Hence, the value of K_{eq} at 25°C is 0.0185512

  • <u>For C:</u>

We are given:

[CoCl_4]^{2-}=0.234625M

[Co(H_2O)_6]^{2+}=0.765375M

[Cl^-]=1.06150M

Putting values in equation 1, we get:

K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886

Hence, the value of K_{eq} at 65°C is 0.2887886

  • <u>For D:</u>

For Endothermic reactions, \Delta H>0, which is positive

For Exothermic reactions, \Delta H, which is negative

To calculate \Delta H of the reaction, we use Van't Hoff's equation, which is:

\ln(\frac{K_{65^oC}}{K_{25^oC}})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{65^oC} = equilibrium constant at 65°C = 0.2887886

K_{25^oC} = equilibrium constant at 25°C = 0.0185512

\Delta H = Enthalpy change of the reaction = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 25^oC=[25+2730]K=298K

T_2 = final temperature = 65^oC=[65+2730]K=338K

Putting values in above equation, we get:

\ln(\frac{0.2887886}{0.0185512})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{298}-\frac{1}{338}]\\\\\Delta H=57471.26J/mol

As, the calculated value of \Delta H>0. Thus, the reaction is endothermic in nature.

4 0
3 years ago
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