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3 years ago

Why does air at 75% relative humidity and 30°C make your skinfeel stickier than air at 75% relative humidity and 10°C

Chemistry

2 answers:

ahrayia [7]3 years ago

8 0

Here are the choices:

Warm air rises while cold air falls.

The warmer air contains less water vapor per unit of volume.

The warmer air contains more water vapor per unit of volume.

Moist air has lower density than dry air does.

The best answer is: <em>The warmer air contains more water vapor per unit of volume. </em>

Valentin [98]3 years ago

4 0

Answer:its more water

Explanation:

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Which organ gives our body the sense to touch, protects it, and helps keep it at the right temperature?

Tomtit [17]

The answer is A) skin

7 0

2 years ago

Read 2 more answers

Be sure to answer all parts. The equilibrium constant (Kp) for the reaction below is 4.40 at 2000. K. H2(g) + CO2(g) ⇌ H2O(g) +

nikklg [1K]

<u>Answer:</u>

<u>For 1:</u> The value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

<u>For 2:</u> The value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

<u>Explanation:</u>

For the given chemical equation:

$H_2(g)+CO_2(g)\rightleftharpoons H_2O(g)+CO(g)$

<u>For 1:</u>

To calculate the $\Delta G$ for given value of equilibrium constant, we use the relation:

$\Delta G=-RT\ln K_p$ .....(1)

where,

$\Delta G$ = ? kJ/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 4.40

Putting values in above equation, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (4.40)\\\\\Delta G=-24636.12J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -24636.12 J/mol = -24.636 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -24.636 kJ/mol

<u>For 2:</u>

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p_{CO}p_{H_2O}}{p_{H_2}p_{CO_2}}$

We are given:

$p_{CO}=1.18atm\\p_{H_2O}=0.66atm\\p_{CO_2}=0.82atm\\p_{H_2}=0.27atm$

Putting values in above equation, we get:

$K_p=\frac{1.18\times 0.66}{0.27\times 0.82}\\\\K_p=3.52$

Now, calculating the value of $\Delta G$ by using equation 1:

R = Gas constant = $8.314J/K mol$

T = temperature = 2000 K

$K_p$ = equilibrium constant in terms of partial pressure = 3.52

Putting values in equation 1, we get:

$\Delta G=-(8.314J/Kmol)\times 2000K\times \ln (3.52)\\\\\Delta G=-20925.68J/mol$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, -20925.68 J/mol = -20.925 kJ/mol

Hence, the value of $\Delta G$ for the chemical equation is -20.925 kJ/mol

3 0

3 years ago

Read 2 more answers

Phosphorus pentachloride reacts with water to form hydrochloric acid and phosphoric acid. How many total moles of acid are forme

klio [65]

Answer:

0,13 moles of acid are produced

Explanation:

The reaction of the problem is:

PCl₅ + 4H₂O → 5HCl + H₃PO₄

Based on the reaction, 1 mole of PCl₅ produces 6 moles of acid (5 moles of HCl + 1 mole of H₃PO₄).

The molecular mass of PCl₅ is:

1P = 30,97g/mol + 5Cl = 5×35,45g/mol = <em>208,24 g/mol</em>

That means 4,5g of PCl₅ are:

4,5g PCl₅×(1mol / 208,24g) = 0,0216 moles of PCl₅. As 1 mole of PCl₅ produces 6 moles of acid, 0,0216 moles of PCl₅ produce:

0,0216 moles PCl₅ × (6 moles acid / mole of PCl₅) =

<em>0,13 moles of acid are produced</em>

I hope it helps!

3 0

2 years ago

How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol

netineya [11]

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

$\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}$

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid