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2 years ago

10

.A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is

1 answer:

Strike441 [17]2 years ago

5 0

Answer:

47.36mL

Explanation:

Using Boyles law equation, which states that:

P1V1 = P2V2

Where;

V1 = initial volume (mL)

V2 = final volume (mL)

P1 = initial pressure (atm)

P2 = final pressure (atm)

Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm

First, we need to convert 152kPa to atm by dividing by 101

1kPa = 0.0099atm

152kPa = 1.505atm

P1V1 = P2V2

1.505 × 25.3 = 0.804 × V2

38.08 = 0.804V2

V2 = 38.08/0.804

V2 = 47.36mL

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If 0.138g of cyclohexene (c6h10) was obtained from 0.240g of cyclohexanol (c6h120), what is the percentage yield of cyclohexene?

tino4ka555 [31]

<span>Given in the question- 1 mole of cyclohexanol = > 1 mole of cyclohexene Molar mass 100.16 g/mol moles of cyclohexanol = .240 / 100.16= 0.002396 moles Molar mass 82.143 g/mol moles of cyclohexene formed @100 % yield = 0.002396 Molar mass 82.143 g/mol mass of cyclohexene @ 100 % = .002396 x 82.143 = 0.197g bur we have .138g so % yield = .138 / .197 = 70.0 % Ans- 70 percentage yield of cyclohexene.</span>

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3 years ago

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liubo4ka [24]

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1 year ago

Read 2 more answers

Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

<u>Answer:</u> The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

<u>Explanation:</u>

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

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Answer:

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Explanation:

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2 years ago

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Explanation:

Thanks google

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