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vazorg [7]
4 years ago
8

What is the coefficient for hydrogen in the balanced equation for the reaction of solid molybdenum(iv) oxide with gaseous hydrog

en to form solid molybdenum and liquid water?
Chemistry
1 answer:
anastassius [24]4 years ago
3 0

The  coefficient  for hydrogen  in  the  balanced  equation   of solid  molybdenum(iV)  oxide  with gaseous  hydrogen  is  2


 Explanation

Coefficient   is defined  to  as  a number  in front   of a chemical formula in a  balanced chemical equation.

   The  reaction   of   molybdenum (iv) oxide  with  gaseous  hydrogen  is  as below,


MoO2  + 2 H2→  Mo  +2 H2O

From   balanced equation above  the coefficient   for  H2  is  2  since  the number in  front of  H2  is 2



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Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
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<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

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Moles of chloride ions in KCl = 0.200 moles

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Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

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