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3 years ago

What are the oxidation numbers of the atoms in this reaction?

Chemistry

2 answers:

SVEN [57.7K]3 years ago

6 0

Answer : The correct options are:

0 for Mg(s) and H in $H_2$

+1 for H in HCl

+2 for Mg in $MgCl_2$

Explanation :

The given chemical reaction is:

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$

Oxidation state of Mg(s) is, 0

Oxidation state of hydrogen in HCl is, (+1)

Oxidation state of chlorine in HCl is, (-1)

Oxidation state of Mg in $MgCl_2$ is, (+2)

Oxidation state of Cl in $MgCl_2$ is, (-1)

Oxidation state of hydrogen in $H_2$ is, 0

The given reaction is a single displacement reaction in which the most reactive metal (Mg) displaces the least reactive element (H) from its compound.

It is also a redox reaction because the oxidation state of Mg changes from 0 to (+2) and shows oxidation and the oxidation state of hydrogen changes from (+1) to 0 and shows reduction.

OLEGan [10]3 years ago

5 0

<span>
all these are correct:
0 for Mg(s) and H in H2

</span><span>+1 for H in HCl

</span><span>-1 for Cl in HCl
</span>
<span>+2 for Mg in MgCl2

only the first statement is wrong</span>

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What is the volume of a weather balloon that has 39 grams of helium with a density 0.017 g/mL?

olchik [2.2K]

Answer:

<h3>The answer is 235.29 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question.

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We have

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We have the final answer as

Hope this helps you

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2 years ago

A mixture of0.161 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L-vessel at 500.0 K, producing a mixture of CO

labwork [276]

Answer:

number of moles of CO2 is 0.054

number of moles of CO is 0.107

number of moles of O2 remaining is 0.01 mole

mole fraction of CO is 0.63

Explanation:

Firstly, we write the equation of reaction;

3C(s) +2O2(g) → CO2(g) +2CO(g)

Now, we proceed.

From the written equation, we can deduce that

3 mol C = 2 mol O2 = 1 mol CO2 = 2 mol CO

No of mol of C reacted = 0.161 mol

limiting reactant according to the question is Carbon

a. no of mol of CO2 formed = 0.161*1/3 = 0.054 moles ( no of moles of CO2 formed is one-third of no of moles of carbon reacted. This is obtainable from their mole ratio 1:3)

b. no of mol of CO formed = 0.161*2/3 = 0.107 mol

c. no of mol of O2 remaining = 0.117 - (0.151*2/3) = 0.117-0.107 = 0.01 mole

d. mole fraction of CO = no of mol of CO/Total number of moles

= 0.107/(0.107+0.054+0.01)

= 0.625730994152 which is approximately 0.63

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3 years ago

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vitfil [10]

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