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3 years ago

Why does air exert pressure?

Chemistry

2 answers:

Licemer1 [7]3 years ago

5 0

Answer: because air has mass.

Reasoning: The Atmosphere has little to do with the concept here.

Wind moving air, why would that resulting in pressure? Does wind push you aside make you exert pressure?

Though increased temperatures can increase pressure. It doesn't affect it fully.

pogonyaev3 years ago

4 0

the correct answer is the first one because air has mass.

when gases molecules hit a surface or wall or in a room, they exert forces on how much particles or how hard their hitting that surface. so the more wild (hot) the molecules are the more pressure there is.

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The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammoni

ValentinkaMS [17]

Answer:

" $6.7\times 10^{-4} \ atm$ " is the right answer.

Explanation:

Given:

Partial pressure of $N_2$ ,

= 0.20 atm

Partial pressure of $H_2$ ,

= 0.15 atm

$K_p = 1.5\times 10^3$ at $400^{\circ} C$

As we know,

⇒ $K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}$

By putting the values, we get

$1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}$

$pNH_3^2 = \frac{0.000675}{1.5\times 10^3}$

$=6.7\times 10^{-4} \ atm$

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2 years ago

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aev [14]

The most concentrated solution is b

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3 years ago

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How many molecules of O2 are required to react with 3.00 moles P4? P4+5O2⟶2P2O5

likoan [24]

Answer:

1400

Explanation:

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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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saul85 [17]

Answer:

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