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cestrela7 [59]
2 years ago
5

A sample of nitrogen gas is collected over water at a temperature of 23.0 C. What is the pressure of the nitrogen gas if atmosph

eric pressure is 785 mm Hg
Chemistry
1 answer:
Whitepunk [10]2 years ago
4 0
We first consider the gases that will be present in that sample.
First, there will be nitrogen, as stated. Second, there will also be water in the form of water vapor. For this, we need the vapor pressure of water at 23.0 °C, which is about 21.0 mmHg. Now, the sum of the vapor pressures of the gases will be equivalent to the total pressure. So the pressure of nitrogen gas is:
785 - 21
= 764 mmHg
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How many grams of potassium chloride are required to make 250 ml of a 0.75 m kcl solution?
DedPeter [7]

Answer:

The solution would need 13.9 g of KCl

Explanation:

0.75 m, means molal concentration

0.75 moles in 1 kg of solvent.

Let's think as an aqueous solution.

250 mL = 250 g, cause water density (1g/mL)

1000 g have 0.75 moles of solute

250 g will have (0.75 . 250)/1000 = 0.1875 moles of KCl

Let's convert that moles in mass (mol . molar mass)

0.1875 m . 74.55 g/m = 13.9 g

7 0
2 years ago
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denis23 [38]
The compounds that are produced upon this combustion reaction would be Carbon Dioxide and water.

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2 years ago
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If the Earth had no tilt, how would that affect our seasons? I will give branliest to first comment
Aleks [24]

Answer:

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6 0
3 years ago
Energy of motion or change in position
Sholpan [36]
<span>Kinetic energy is the energy at move or at motion. The kinetic energy of objects depend on its mass (m) and its </span>velocity (v)<span>. The mass of an object can be measured in kilograms while its mass can be measured in meters per second. The objects' energy at motion is very dependent to its mass and velocity therefore, the formula in finding the kinetic energy of an object is </span>

K = 1/2 x mass (m) x velocity (m/s)2

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8 0
3 years ago
For the gas phase decomposition of t-butyl propionate,C2H5COOC(CH3)3 (CH3)2C=CH2 + C2H5COOHthe rate constant at 540 K is 8.90×10
kvv77 [185]

Answer:

The activation energy is 164.02 kJ/mol

Explanation:

Log (k2/k1) = Ea/2.303R × [1/T1 - 1/T2]

k1 = 8.9×10^-4 s^-1

k2 = 9.83×10^-3 s^-1

R = 8.314 J/mol.K

T1 = 540 K

T2 = 578 K

Log (9.83×10^-3/8.9×10^-4) = Ea/2.303×8.314 × [1/540 - 1/578]

1.043 = 6.359×10^-6Ea

Ea = 1.043/6.359×10^-6 = 164020 J/mol = 164020/1000 = 164.02 kJ/mol

6 0
3 years ago
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