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Free_Kalibri [48]
3 years ago
8

Which of the following actions would cause the volume of the chamber below to increase to 36.0 L?

Chemistry
1 answer:
Orlov [11]3 years ago
5 0

Answer:

Option C. Triple the number of moles

Explanation:

From the ideal gas equation:

PV = nRT

Where:

P is the pressure

V is the volume

n is the number of mole

R is the gas constant

T is the absolute temperature.

Making V the subject of the above equation, we have:

PV = nRT

Divide both side by P

V = nRT / P

Thus, we can say that the volume (V) is directly proportional to both the number of mole (n) and absolute temperature (T) and inversely proportional to the pressure (P). This implies that and increase in either the number of mole, the absolute temperature and a decrease in the presence will cause the volume to increase.

Thus, the correct option is option C triple the number of moles. This can further be seen as illustrated below:

Initial volume (V1) = 12 L

Initial mole (n1) = 0.5 mole

Final mole (n2) = triple the initial mole = 3 × 0.5 = 1.5 mole

Final volume (V2) =?

From:

V = nRT / P, keeping T and P constant, we have:

V1/n1 = V2/n2

12/0.5 = V2/1.5

24 = V2/1.5

Cross multiply

V2 = 24 × 1.5

V2 = 36 L.

Thus Option C gives the correct answer to the question.

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Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>

The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:

  • molar mass = density × molar gas constant × temperature/pressure

Molar gas constant, R = R = 0.082 L.atm/mol/K.

Temperature = 150 °C = 423 K

Pressure = 785 torr = 1.033 atm

density = 4.93 g/L

molar mass of gas = 4.93 × 0.082 × 423/1.033

molar mass of gas = 165.5 g/mol

Then, molecular weight of the gas = 165.5 amu

Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

Learn more about molar mass of a gas at: brainly.com/question/26215522

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2 years ago
How do scientist communicate the results of an investigation?​
maks197457 [2]

Answer:

scientists often communicate their research results in three general ways:

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2) Two is to present their results at national and international conferences where other scientists can listen to presentations

Explanation:

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the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -&gt; 2AlCl3 based on this equation how many molecules of chlorin
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45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = \frac{30 * 3}{2}\\ = 45 molecules of Cl₂.

Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

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2 years ago
Which of these compounds is a proton acceptor ?
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Answer:

Sodium Hydroxide

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3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
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Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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