Calculate the volume of each of the following gases at STP 7.6 mol Ar 0.44 mol C2,H3

Answers

Answer 1

Answer: For Ar :

1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L

x = 7.6 * 22.4

x = 170.24 L
-----------------------------------------------------------------
For C2H3:

1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L

y = 0.44 * 22.4

y = 9.856 L

hope this helps !.

Answer 2

Answer:

The volume of 0.44 moles of ethylene gas at STP is approximately 10.33 liters.

To calculate the volume of gases at Standard Temperature and Pressure (STP), we can use the ideal gas law:

PV = nRT

Where:

P = Pressure (at STP, it's 1 atmosphere, or 1 atm)

V = Volume (in liters)

n = Number of moles

R = Ideal gas constant (0.0821 L·atm/mol·K)

T = Temperature (at STP, it's 273.15 K)

Let's calculate the volume for each gas:

For 7.6 moles of Ar (argon):

P = 1 atm

n = 7.6 moles

R = 0.0821 L·atm/mol·K

T = 273.15 K

Now, plug these values into the ideal gas law:

V = (nRT) / P

V = (7.6 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 172.75 liters

So, the volume of 7.6 moles of argon gas at STP is approximately 172.75 liters.

For 0.44 moles of C2H3 (ethylene):

P = 1 atm

n = 0.44 moles

R = 0.0821 L·atm/mol·K

T = 273.15 K

Now, plug these values into the ideal gas law:

V = (nRT) / P

V = (0.44 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 10.33 liters

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Answers

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A student made a copper bracelet by hammering a small copper bar into the desired shape. The bracelet has a mass of 30.1 grams and was at a temperature of 21°C in the classroom. After the student wore the bracelet, the bracelet reached a temperature of 33°C. Later, the student removed the bracelet and placed it on a desk at home, where it cooled from 33°C to 19°C. The specific heat capacity of copper is 0.385 J/g•K.69 Explain, in terms of heat flow, the change in the temperature of the bracelet when the student wore the bracelet. [1]

70 Determine the number of moles of copper in the bracelet. [1]

71 Show a numerical setup for calculating the amount of heat released by the bracelet as it cooled on the desk. [1]

72 Explain, in terms of chemical activity, why copper is a better choice than iron to make the bracelet. [1]

Answers

69. The student's skin was obviously warmer than the initial temperature of the bracelet. According to the second law of thermodynamics, heat transfers from a higher temperature to a lower temperature. Hence, heat transfers from the student's skin to the bracelet warming up the bracelet.

70. To determine number of moles of copper:

Given mass=30.1 grams

Molar mass= 63.55 g/mol

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\n\n \text{Number of moles}=(30.1)/(63.55) =0.4736 \text { moles}$

Thus, the number of moles of copper is 0.4736.

71. Calculating the amount of heat released:

ΔH( heat released) = m Cp(ΔT)

= 30.1g(0.385 J/g. K)(33-19)K

ΔH= 162.23J

Thus, the amount of heat released is 162.23 J.

72. Copper is chemically less active than iron.

Copperdoes notdisplace iron from its solution since it is less reactive than iron. A more reactive metal displaces a less reactive metal from its solution.

Learn more:

brainly.com/question/25548593

69. The student's skin was obviously warmer than the initial temperature of the bracelet. According to the second law of thermodynamics, heat transfers from a higher temperature to a lower temperature hence heat transfers from the student's skin to the bracelet warming up the bracelet.

70. moles of copper=mass of copper/molar mass of copper
=30.1g/(63.55g/mol)= 0.4736 moles Cu

71. Delta H( heat released) = mCp(deltaT) = 30.1g(0.385 J/gK)(33-19)K = 162.23J

72. Copper is chemically less active than iron.

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.4-L bulb, then filled it with the gas at 1.00 atm and 22.0 ∘C and weighed it again. The difference in mass was 5.1 g . Identify the gas.

Answers

Answer : The diatomic gas is nitrogen gas, N₂.

Explanation :

First we have to calculate the moles of gas.

Using ideal gas equation:

$PV=nRT$

where,

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T = Temperature of gas = $22.0^oC=273+22.0=295.0K$

Putting values in above equation, we get:

$1.00atm* 4.4L=n* (0.0821L.atm/mol.K)* 295.0K$

$n=0.1817mol$

Now we have to calculate the molar mass of gas.

$\text{Molar mass of gas}=\frac{\text{Given mass of gas}}{\text{Moles of gas}}$

$\text{Molar mass of gas}=(5.1g)/(0.1817mol)=28.07g/mol$

As we are given that the gas is diatomic X₂.

As, 2 atoms of gas X has mass = 28.07 g/mol

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