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erastovalidia [21]
3 years ago
5

Please and thank you!

Chemistry
1 answer:
Valentin [98]3 years ago
3 0

Answer:

7.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.

Thus, 7.5 moles of O₂ were obtained from the reaction.

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What is the molar mass of sodium phosphate, Na3PO4?
baherus [9]

Answer:

164g

Explanation:

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8 0
3 years ago
How many ml of a 2.0 m nabr solution are needed to make 200.0 ml of 0.50 m nabr?
gogolik [260]

To solve this we use the equation,

 

<span> M1V1 = M2V2</span>

 

<span> where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.</span>

 

<span>2.0 M x V1 = 0.50 M x 200 mL</span>

<span>V1 = 50 mL needed</span>

4 0
3 years ago
If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<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)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

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