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lara31 [8.8K]
2 years ago
6

A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL

volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (7.0M) stock nitric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated nitric acid
Chemistry
1 answer:
d1i1m1o1n [39]2 years ago
6 0

<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

pH = 0.70

Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL

Hence, the volume of concentrated solution required is 9.95 mL

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AlexFokin [52]

Answer:

2C₂H₆ +  [7]O₂     →      [4]CO₂ + [6]H₂O

Explanation:

Chemical equation:

C₂H₆ +  O₂     →      CO₂ + H₂O

Balanced chemical equation:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Step 1:

2C₂H₆ +  O₂     →      CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C = 1

H = 12                                    H = 2

O = 2                                     O = 3

Step 2:

2C₂H₆ +  O₂     →      4CO₂ + H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 2

O = 2                                     O = 9

Step 3:

2C₂H₆ +  O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 2                                     O = 14

Step 4:

2C₂H₆ +  7O₂     →      4CO₂ + 6H₂O

Left hand side                      Right hand side

C = 4                                     C =  4

H = 12                                    H = 12

O = 14                                     O = 14

3 0
3 years ago
Which of the following concerning the effects of temperature and pressure on solubility is/are correct?
just olya [345]

Answer:

None of the options are correct.

Explanation:

1) when the temperature of the solution is increased the solubility of the gas in the liquid decreases , hence option 1 is incorrect.

2)The heat released by the dissolution of an ionic compound in water is heat of hydration of the compound and is independent of the initial temperature of the solution.

3) The solubility of a liquid in water is not affected significantly by the pressure changes in the system as gases only have a significant cahne in solubility with change in pressure.

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3 years ago
What is the density of an unknown substance that has a mass of 30 g and a volume of 10mL? a. 300 g/mL b. 30 g/mL c.3 g/mL d.1 g/
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Answer:

the answer is 3g/cm3 I am pretty sure

6 0
2 years ago
What is the empirical formula for a compound composed of 0.0683 mol of carbon ( C ), 0.0341 mol of hydrogen ( H ), and 0.1024 mo
liubo4ka [24]

The empirical formula for a compound composed of 0.0683 mol of carbon ( C ), 0.0341 mol of hydrogen ( H ), and 0.1024 mol of nitrogen ( N ) is  C_2HN_3.

<h3>What is the empirical formula?</h3>

An empirical formula tells us the relative ratios of different atoms in a compound.

Given data:

Moles of carbon = 0.0683 mol

Moles of hydrogen = 0.0341 mol

Moles of nitrogen = 0.1024 mol

Dividing each mole using the smallest number that is divided by 0.0341 moles.

We get:

Carbon= 2

Hydrogen=1

Nitrogen=3

The empirical formula for a compound isC_2HN_3.

Learn more about empirical formula here:

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4 0
1 year ago
Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume
Kisachek [45]

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

Explanation:

the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

Na2S2O3 = 0.05 * 0.005 = 0.00025 moles

KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

3 0
3 years ago
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