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Eddi Din [679]
3 years ago
12

Calculate the mass of NaCl needed to make 40% solution from 3kg of water ? Could you explain step by step please :-)

Chemistry
1 answer:
earnstyle [38]3 years ago
3 0
Ok so 40 percent of 3000 grams which is 3 kg is 1200 grams 
So you would need 1.2kg or 1200 grams to make a 40% solution
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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
Calculate the molar concentration of the Br⁻ ions in 0.065 M MgBr2(aq).
steposvetlana [31]
MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
MgBr2(s) --> Mg+(aq) + 2 Br⁻(aq)
            [Br⁻] = 0.065 mol MgBr2/1L × 2 mol Br⁻ / 1 mol MgBr2 = 0.13 M
The answer to this question is [Br⁻] = 0.13 M
4 0
3 years ago
All of the following are true of bases except
Oksana_A [137]
All of these are correct except the first option, as Arrhenius bases increase the concentration of hydroxide ions.
4 0
3 years ago
Read 2 more answers
The molarity of concentrated acetic acid is 17.45 M. When acetic acid is diluted in water it is commonly called vinegar with a m
Sladkaya [172]

M1 = 17.45 M

M2 = 0.83 M

V2 = 250 ml

M1. V1= M2. V2

V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml

7 0
2 years ago
A buffered solution has a pH of 7.5. What would happen to the pH if a small
musickatia [10]

Answer:

Dear user,

Answer to your query is provided below

When small amount of acid was added to buffered solution, pH will change very less.

Explanation:

Buffer solution resists change in ph on adding small amount of acid or base but when we calculate the value of buffer capacity we take the change in ph when we add acid or base to 1 lit solution of buffer.This contradicts the definition of buffer solution.

5 0
3 years ago
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