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Natali5045456 [20]

3 years ago

10

For 520.0 ml of a buffer solution that is 0.175 m in hc2h3o2 and 0.165 m in nac2h3o2, calculate the initial ph and the final ph

after adding 0.020 mol of hcl.

1 answer:

alisha [4.7K]3 years ago

3 0

KWIJRJHO0907 X 023KO

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SSSSS [86.1K]

Answer:

pH is the number given to a substance of how powerful it is before 7 or beyond 7.

the solution is an acid.

Explanation:

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3 years ago

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What compound is Rb2CrO4

Nimfa-mama [501]

Rubidium Chromate is the compound

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3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

<u>The reaction </u>

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) <u>Limiting reagent</u>: $NO$

3) <u>Ammount of excess reagent</u>:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

8 0

3 years ago

Determine how many carbon dioxide (CO2) molecules are produced if 8.45 x 1023 molecules of water (H2O) are produced 2 C2H6 (g) +

Andreas93 [3]

Ill get back to you on this

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3 years ago

How can the rate constant be determined from the rate law?

Vinil7 [7]

Answer:

B. The rate constant is the reaction rate divided by the concentration

terms.

Explanation:

The rate constant can be determined from the rate law because it is the reaction rate divided by the concentration terms. I hope I could help! :)

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2 years ago