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2 years ago

S + 6 HNO3 → H2SO4 + 6 NO2 + 2 H2O

Chemistry

1 answer:

mote1985 [20]2 years ago

4 0

Answer:

0.9grams

Explanation:

IF based on stoichiometry HNO3 to H2O is 6:2

Use 9 grams to find the moles of HNO3

9 grams/63g/mol=0.143 moles of HNO3

HNO3:H2O

6:2

0.143*2/6=0.048

H2O moles is 0.048moles

Mass of water =0.048moles*18g/mol=0.864g

To the nearest tenth=0.9grams.

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3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

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3 years ago

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Aleks04 [339]

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3 years ago

If 1.546 g of copper was used by a student at the start of the lab, and 0.732 g of copper were obtained at

mash [69]

Answer: Percent recovery is 47.34 %

Explanation:

Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

${\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100$

Putting in the values we get:

${\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%$

Therefore, the percent recovery is 47.34 %

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3 years ago