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liq [111]
3 years ago
15

A rigid vessel contains water at 0.15 MPa (at 25 °C). The vessel is heated to the critical point of water. Calculate the fractio

n of mass of liquid at 25°C.
Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

0.8078 Kg

Explanation:

Pressure of water = 0.15 MPa = 1.5 bar

At critical point of water ,temperature = 647 K=374°C

From the ideal gas equation

P×V= m×R×T

Let us assume volume = 1 m^3

1.5 x 105 x 1 = m x 287 x 647

m= 0.8078 kg

the fraction of mass of liquid at 25°C.

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2As2S3 + 9O2 = 2As2O3 + 6SO2

Explanation:

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For the wild type (unmutated) enzyme, you measure a rate of p-nitrophenol release by the change in absorbance at 405 nm (for the
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Answer:

1.2x10⁻⁵M = Concentration of the product released

Explanation:

Lambert-Beer's law states the absorbance of a solution is directly proportional to its concentration. The equation is:

A = E*b*C

<em>Where A is the absotbance of the solution: 0.216</em>

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<em />

Replacing:

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<h3>1.2x10⁻⁵M = Concentration of the product released</h3>
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What property is used to calculate the ph of a solution
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A removal of the predator populations from an ecosystem would most likely result in
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The increase of prey

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

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3 years ago
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