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3 years ago

7

A 0.211 g sample of carbon dioxide, CO2, has a volume of 560 mL and a pressure of 429 mmHg whats the temperature of the gas in K

elvin

1 answer:

SCORPION-xisa [38]3 years ago

5 0

The temperature of the gas sample is 813 K.

<u>Explanation:</u>

We have to use the ideal gas equation to find the temperature of the gas sample.

The ideal gas equation is PV = nRT

Pressure, P = 429 mm Hg = 0.56 atm

Volume, V = 560 mL = 0.56 L

R = gas constant = 0.08205 L atm mol⁻¹K⁻¹

Mass = 0.211 g

Molar mass of carbon di oxide = 44.01 g / mol

Moles, n = $$\frac{given mass}{molar mass} = \frac{0.211 g}{44.01 g/mol}$

= 0.0047 mol

Now, we have to plugin the above values in the above equation, we will get the temperature as,

$$T= \frac{PV}{nR}$

T = $$\frac{0.56 \times 0.56}{0.08205 \times 0.0047}$

= 813 K

So the temperature of the gas sample is 813 K.

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A gas occupies 56 L at 73°C. What volume will the gas occupy if the temp. cools to 0°C?

vova2212 [387]

Answer:

44.2 L

Explanation:

Use Charles Law:

$\frac{V1}{T1} =\frac{V2}{T2}$

We have all the values except for V₂; this is what we're solving for. Input the values:

$\frac{56 L}{346K} =\frac{V2}{273K}$ - make sure that your temperature is in Kelvin

From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

$\frac{56*273}{346} = V2$

Therefore, V₂ = 44.2 L

It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.

7 0

2 years ago

When salt is dissolved in water, what happens to the water

choli [55]

Answer:

<em>When salt is dissolved in water</em>, many physical properties change, among them the so called colligative properties:

The vapor pressure of water decreases,
The boiling point increases,
The freezing point decreases, and
Osmotic pressure appears.

Explanation:

Colligative properties are the physical properties of the solvents whose change is determined by the number of particles (moles or ions) of the solute added.

The colligative properties are: vapor pressure, boiling point, freezing point, and osmotic pressure.

<u>Vapor pressure</u>:

The vapor pressure is the pressure exerted by the vapor of a lquid over its surface, in a closed vessel.

The vapor pressure increases when a solute is added, because the presence of the solute causes less solvent molecules to be near the surface ready to escape to the vapor phase, which means that the vapor pressure is lower.

<u>Boiling point</u>:

The boiling point is the temperature at which the vapor pressure of the liquid equals the atmospheric pressure. Since we have seen that the vapor pressure of water decreases when a solute occupies part of the surface, now more temperature will be required for the water molecules reach the atmospheric pressure. So, the boiling point increases when salt is dissolved in water.

<u>Freezing point</u>:

The freezing point is the temperarute at which the vapor pressure of the liquid and the solid are equal. Since, the vapor pressure of water with salt is lower than that of the pure water, the vapor pressure of the liquid and solid with salt will be equal at a lower temperature. Hence, the freezing point is lower (decreases).

<u>Osmotic pressure</u>:

Osmotic pressure is the additional pressure that must be exerted over a solution to make that the vapor pressure of the solvent in the solution equals the vapor pressure of the pure solvent. This additional pressure is proportional to the concentration of the solute: the higher the salt concentration the higher the osmotic pressure.

6 0

3 years ago

Consider this question: What is the molarity of HCL if 35.23 mL of a solution of HCL contains 0.3366 g of HCL?

Crazy boy [7]

<u>Answer:</u> The molarity of HCl solution is 0.262 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of HCl = 0.3366 g

Molar mass of HCl = 36.5 g/mol

Volume of the solution = 35.23 mL

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.3366g\times 1000}{36.5g/mol\times 35.23mL}\\\\\text{Molarity of HCl}=0.262M$

Hence, the molarity of HCl solution is 0.262 M.

6 0

2 years ago

A flame is the result of a chemical reaction that releases energy. Do you think a similar kind of chemical reaction is happening

almond37 [142]

Answer:

Chemical reactions that take place inside living things are called biochemical reactions. Explanation:The sum of all the biochemical reactions in an organism is referred to as metabolism. Metabolism includes both exothermic (heat-releasing) chemical reactions and endothermic (heat-absorbing) chemical reactions.

4 0

2 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago