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stepan [7]
3 years ago
5

If you have 1kg of carbon and 1 kg of gold, they will weigh the same.

Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
8 0
This would be be true
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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24
Setler79 [48]

<u>Answer:</u>

<u>For Part A:</u> The partial pressure of Helium is 218 mmHg.

<u>For Part B:</u> The mass of helium gas is 0.504 g.

<u>Explanation:</u>

  • <u>For Part A:</u>

We are given:

p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg

To calculate the partial pressure of helium, we use the formula:

P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}

Putting values in above equation, we get:

745=245+119+163+p_{He}\\p_{He}=218mmHg

Hence, the partial pressure of Helium is 218 mmHg.

  • <u>For Part B:</u>

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

PV=\frac{m}{M}RT

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g

Hence, the mass of helium gas is 0.504 g.

6 0
3 years ago
Propane (c3h8) is burned in oxygen to produce carbon dioxide and water. the heat of combustion of propane is -2012 kj/mole. how
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C_{3} H_{8} + 5 O_{2} ---\ \textgreater \  3CO_{2}  +4H_{2}O    (-2012 \frac{kJ}{mol} )&#10;&#10;&#10;3 mol                10 mol&#10;&#10;&#10;C_{3}H_{8} is /excess /reactant&#10;&#10;because 3 mol propane require 15 mol oxygen (by reaction)&#10;&#10;5 mol oxygen ---1 mol propane, so&#10;&#10;10 mol oxygen ---2 mol propane&#10;&#10;Only 2 mole propane will be burned,&#10;&#10;so &#10;&#10;2012 ( kJ/mol)*2 mol =4024 KJ heat will be given off&#10;&#10;Correct answer is number 4.&#10;&#10;&#10;
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3 years ago
Why is it important for you to learn where the safety equipment is located in the lab?
Mariana [72]
It is really very important to know where the safety equipments are placed in lab because when we know it then accident wont occur in lab . it will also helps us to feel secure while doing any kinds of experiment
6 0
3 years ago
Please help! BRAINLIEST to right answerrrr
zloy xaker [14]

Answer: from the hotter surface to the colder one

Explanation:

8 0
3 years ago
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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
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