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4 years ago

5

A student is in the lab and combines two unknown substances ( which you should never do because it is not safe). one is a white

solid, and the other one is clear liquid. after the substances are mixed, you observe...
1- bubbles forming,
2- dementia turns a dark purple color, and
3-the white solid substance dissolves into the liquid.

did a physical change or a chemical change happen? which of the observation would you use as evidence in why?

1 answer:

dem82 [27]4 years ago

7 0

Answer:

Chemical change

Explanation:

If you were to mix two unknown substances and a reaction as such was to happen it would most likely be a Chemical change as it had a reaction with the other substance and created a chemical reaction.

Sorry for any spelling errors English is my second language.

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3 years ago

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[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.

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