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Zinaida [17]
3 years ago
6

Enter your answer in the provided box.

Chemistry
1 answer:
olga55 [171]3 years ago
4 0

Answer:

10.5L

Explanation:

The volume in this question can be calculated by using the formula for gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in this question:

P = 6.18 atm

V = ?

n = 2.35 moles

R = 0.0821 Latm/molK

T = 63°C = 63 + 273 = 336K

Using PV = nRT

V = nRT/P

V = 2.35 × 0.0821 × 336/ 6.18

V = 64.83/6.18

V = 10.49

V = 10.5L

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Hydrogen is a possible future fuel. However, elemental hydrogen is rare, so it must be obtained from a hydrogen- containing comp
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Answer:

1.1 × 10² g

Explanation:

First, we will convert 1.0 L to cubic centimeters.

1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³

The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:

1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g

1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:

1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g

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Use the following reaction: C4H9OH NaBr H2SO4 C4H9Br NaHSO4 H2O If 15.0 g of C4H9OH react with 22.4 g of NaBr and 32.7 g of H2SO
ohaa [14]

Answer:

Percent yield = 61.58%

Explanation:

C4H9OH + NaBr + H2SO4 --> C4H9Br + NaHSO4 + H2O

From the reaction;

1  mol of C4H9OH reacts with 1 mol of NaBr and 1 mol of H2SO4 to form 1 mol of  C4H9Br

We have to obtain the limiting reagent. We do this by calculatig the mols of each reactant in the reaction;

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C4H9OH; 15 / 74g/mol = 0.2027

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H2SO4; 32.7 / 98 = 0.3336

In this case, the limiting reagent is C4H9OH, it determines the amount of product formed.

From the stoichometry of the reaction; one mole of C4H9OH forms one mole of C4H9Br. This means 0.2027 mol of C4H9Br was formed.

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