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Oxana [17]
2 years ago
8

The question in the picture, I really a correct answer, no cheap answers​

Chemistry
1 answer:
Delvig [45]2 years ago
6 0

Answer:

94.325 g

Explanation:

We'll begin by converting 350 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

350 mL = 350 mL × 1 L /1000 mL

350 mL = 0.35 L

Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:

Volume = 0.35 L

Molarity of KC₂H₃O₂ = 2.75 M

Mole of KC₂H₃O₂ =?

Molarity = mole /Volume

2.75 = Mole of KC₂H₃O₂ / 0.35

Cross multiply

Mole of KC₂H₃O₂ = 2.75 × 0.35

Mole of KC₂H₃O₂ = 0.9625 mole

Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:

Mole of KC₂H₃O₂ = 0.9625 mole

Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)

= 39 + 24 + 3 + 32

= 98 g/mol

Mass of KC₂H₃O₂ =?

Mass = mole × molar mass

Mass of KC₂H₃O₂ = 0.9625 × 98

Mass of KC₂H₃O₂ = 94.325 g

Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g

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If you use 1 mole of NaOH, how much NaAl(OH)4 is produced
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Answer:
             81.97 g of NaAl(OH)₄

Solution:
              The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,

                 2 NaOH  +  2 Al  +  6 H₂O    →    2 NaAl(OH)₄  +  3 H₂

According to this equation,

          2 Moles of NaOH produces  =  163.94 g (2 mole) of NaAl(OH)₄
So,
       1 Mole of NaOH will produce  =  X g of NaAl(OH)₄

Solving for X,
                     X  =  (1 mol × 163.94 g) ÷ 2 mol

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A mixture of carbon dioxide and hydrogen gases contains carbon dioxide at a partial pressure of 401 mm Hg and hydrogen at a part
AVprozaik [17]

Answer:

X_{H_2}=\0.3584

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Explanation:

According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.

Partial pressure of carbon dioxide = 401 mmHg

Partial pressure of hydrogen gas = 224 mmHg

Total pressure,P = sum of the partial pressure  of the gases = 401 + 224 mmHg = 625 mmHg

Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.

So,

P_{CO_2}=X_{CO_2}\times P

X_{CO_2}=\frac{P_{CO_2}}{P}

X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}

X_{CO_2}=\0.6416

Similarly,

X_{H_2}=\frac{P_{H_2}}{P}

X_{H_2}=\frac{224\ mmHg}{625\ mmHg}

X_{H_2}=\0.3584

5 0
3 years ago
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