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mina [271]
3 years ago
6

What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?

Chemistry
1 answer:
Ymorist [56]3 years ago
5 0

The   molarity  of (HNO₃) that was used  if 2.00 L must  be used   to prepare 4.5 L  of a 0.25M HNO₃ solution is   0.563 M


 <u><em>calculation</em></u>

  This is calculated  usind  M₁V₁=M₂V₂  formula

where,

         M₁(  molarity ₁) = ?

         V₁( volume ₁) = 2.00 L

        M₁ (molarity ₂) = 0.25M

         V₂( volume₂) = 4.5 L

make M₁ the subject  of the formula by  diving both side of the formula  by V₁

   M₁  is therefore = M₂V₂/V₁

M₁ =[ (0.25 M  x 4.5 L) / 2.00 L ]  =0.563 M

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How many significant figures<br> are in this number?<br> 107.051
Anestetic [448]

Answer:

there are 6 significant figures in 107.051

7 0
4 years ago
Read 2 more answers
9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1
WITCHER [35]

Answer:

Keq=11.5

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)

We can write the law of mass action as:

Keq=\frac{[CH_3OH]}{[CO][H_2]^2}

That in terms of the change x due to the reaction extent we can write:

Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}

Nevertheless, for the carbon monoxide, we can directly compute x as shown below:

[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\

[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\

[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\

x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M

Finally, we can compute the equilibrium constant:

Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5

Best regards.

3 0
4 years ago
Please and thank you
Virty [35]

3.8mL of 0.42 phosphoric acid is required.

Reaction

2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl

moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol

moles of H3Po4

= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2

= 0.00106 mol

V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL

V of H3PO4=3.8mL

To know more about calculation in milliliters refer to:-

brainly.com/question/23276655

#SPJ10

4 0
2 years ago
1. How does cellular respiration add carbon to the<br> atmosphere?
asambeis [7]
Glucose and oxygen are changed into energy and carbon dioxide during cellular respiration and that’s how carbon dioxide is released in the air.
8 0
3 years ago
How does an increase in reactant concentration affect the rate of reaction?
Leto [7]

Answer:

increase the rate of reaction.

Explanation:

7 0
3 years ago
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