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2 years ago

In the unbalanced equation given below, what is the element that is gaining electrons?

Chemistry

2 answers:

DochEvi [55]2 years ago

5 0

Answer:

Mn is gaining electrons.

Explanation:

Let us consider the oxidation state of each element in the given equation.

a) H in HCl : +1

b) Mn in MnO₂ : +4

c) Mn in MnCl₂ : +2

d) H in H₂O : +1

e) Cl in Cl₂ : 0

There is a decrease in oxidation number of Mn from MnO₂ to MnCl₂.

Thus Mn is gaining electrons in the reaction.

Mn is undergoing reduction in the process and chlorine is undergoing oxidation.

viktelen [127]2 years ago

4 0

Hey there!:

HCl + MnO2 → MnCl2 + H2O + Cl2

* in HCl the oxidation state of Cl is -1 .

* on the product side the oxidation state is 0 .

* therefore Cl gains electrons .

* in MnO2 the oxidation state of Mn is +4

* in MnCl2 the oxidation state of Mn is +2

Therefore Mn loses electrons

Answer A

Hope That helps!

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A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adeno

Archy [21]

Answer:

ΔG° of reaction = -47.3 x $10^{3}$ J/mol

Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

K = 1.47 x $10^{-3}$ M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5 $10^{3}$ ) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

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5 0

2 years ago

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

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3 years ago

How can miscible liquids that different boiling points be separated

Afina-wow [57]

Answer:

By boiling and further condensing the liquid with the lowest boiling point.

Explanation:

Hello there!

In this case, according to the attached diagram, it turns out possible for us to infer that the mechanism whereby miscible liquids with different boiling points are separated is distillation, because the flask is heated until the boiling point of the liquid with the lowest value, in order to boil it and subsequently condense it, whereas the liquid with the highest boiling point remains in the flask; and therefore, the two liquids are separated.

Regards!

3 0

2 years ago