All categories

3 years ago

Why cant flame color determine the identity of an ion

1 answer:

6 0

The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color the salt turns the flame of the bunsen burner

You might be interested in

Is the slight bending of light as it passes around the edge of an object.

Snezhnost [94]

The answer to the question Retraction. Its because the light goes around it

7 0

4 years ago

When matter changes from a solid to a liquid, the process is called __________

Anna [14]

Answer: melting

Explanation:

The process in which a solid changes to a liquid is called melting. The melting point is the temperature at which a solid changes to a liquid. For a given type of matter, the melting point is the same as the freezing point.

6 0

3 years ago

What is a stream of charged protons and electrons

Law Incorporation [45]

Answer:

It is called the solar wind thx for asking brainly.com

Explanation:

5 0

4 years ago

[Co(H2O)6]2+(aq) + 4Cl-(aq) ⇌ [CoCl4]2-(aq) + 6H2O(l)

pav-90 [236]

Answer:

For A: The expression for $K_{eq}$ is given below.

For B: The value of $K_{eq}$ at 25°C is 0.0185512

For C: The value of $K_{eq}$ at 65°C is 0.2887886

For D: The reaction is endothermic in nature.

Explanation:

For A:

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.

For B:

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

For C:

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

For D:

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

Explain why children should not be forced to choose sides in a divorce situation, provide two responses

ra1l [238]

Children should not be put into the situations such as the one depicted in this given item. They are still to young to weigh in as to which parent will they will be happier with. Also, they will be very much burdened with the idea of splitting parents disallowing them to focus on other things currently going on.

5 0

3 years ago