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zhenek [66]
3 years ago
14

7.24 miles of Ag(PO4)2

Chemistry
1 answer:
Vladimir [108]3 years ago
7 0
7.42 moles of Ag(PO4)2 x 297.8109g / 1 moles = 2210g Ag(PO4)2
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How many moles of H2O will be formed from the reaction of 80 g of
Roman55 [17]

Answer: 2 moles of H_2O will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} NaOH=\frac{80g}{40g/mol}=2moles

The balanced chemical equation is:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to stoichiometry :

2 moles of NaOH give = 2 moles of H_2O

Thus 2 moles of NaOH give =\frac{2}{2}\times 2=2moles of H_2O

Thus 2 moles of H_2O will be formed.

4 0
3 years ago
What is ph level of acid rain​
KonstantinChe [14]

Answer:

Normal, clean rain has a pH value of between 5.0 and 5.5, which is slightly acidic. However, when rain combines with sulfur dioxide or nitrogen oxides—produced from power plants and automobiles—the rain becomes much more acidic. Typical acid rain has a pH value of 4.0.

Explanation:

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8 0
2 years ago
Describe why corrosion is a natural process
Sladkaya [172]

<u>Answer :</u>

  • <u>Answer :because it happens due to moisture and oxygen </u>

6 0
3 years ago
Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3
Pachacha [2.7K]

<u>Answer:</u> The volume of barium chlorate is 195.65 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL

Hence, the volume of barium chlorate is 195.65 mL

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te
klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

4 0
3 years ago
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