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trapecia [35]
3 years ago
6

What is the volume, in liters, occupied by 1.73 moles of N2 gas at 0.992 atm pressure and a temperature of 75º C? (R value- 0.08

206)
Chemistry
1 answer:
taurus [48]3 years ago
7 0

Volume of the nitrogen gas = 49.8 L

<u>Explanation:</u>

It is given that the pressure, number of moles and temperature of nitrogen gas, and gas constant value being constant and it is taken as 0.08206 L atm mol⁻¹K⁻¹.

Temperature = T = 75°C = 75 + 273 = 348 K

Pressure = P = 0.992 atm

Number of moles = n = 1.73 moles

We have to use the ideal gas equation, PV = nRT, and rearranging the equation to get Volume in litres.

V = $\frac{nRT}{P}

 = $\frac{1.73\times 0.08206\times348}{0.992}

= 49.8 L

So the volume of Nitrogen gas = 49.8 L

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What is the mass of 5.31×1021 platinum atoms?
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The heat of combustion (∆H) for an unknown hydrocarbon is -8.21 kJ/mol. If 0.424 mol of the hydrocarbon is burned in a bomb calo
klio [65]

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

The heat of combustion (∆Hc) for an unknown hydrocarbon is -8.21 kJ/mol. The heat released by the combustion of 0.424 moles of the hydrocarbon is:

Qc = 0.424 mol \times \frac{(-8.21kJ)}{mol} = -3.48 kJ

According to the law of conservation of energy, the sum of the heat released by the combustion (Qc) and the heat absorbed by the bomb calorimeter (Qb) is zero.

Qc + Qb = 0\\\\Qb = -Qc = 3.48 kJ

Given the heat absorbed by the bomb calorimeter (Qb) and the heat capacity of the bomb calorimeter (C), we can calculate the temperature change (ΔT) using the following expression.

Qb = C \times \Delta T\\\\\Delta T = \frac{Qb}{C} = \frac{3.48 kJ}{1.12 kJ/\° C } = 3.10 \° C

When 0.424 moles of an unknown hydrocarbon (∆Hc = -8.21 kJ/mol) is burned in a bomb calorimeter (C = 1.12 kJ/°C), the change in the temperature is 3.10 °C.

Learn more: brainly.com/question/24245395

8 0
3 years ago
One source of aluminum metal is alumina, Al2O3. a. Determine the percentage composition of Al in alumina. b. How many pounds of
Gnom [1K]

Answer:

sorry i cant give the answer but you can gi end check in answer sheet of this becoz i had same question in exam in chemistry so i revised then i checked the answer if you want to check answer go to www.coachscotchemistry.com there you xan find the answer

3 0
3 years ago
A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
4 years ago
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