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ZanzabumX [31]
3 years ago
12

A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

ined 3.221 g. It is again evacuated and then filled with a gas of unknown molar mass. When reweighed, the flask is found to have gained 8.107 g. Part A To estimate the molar mass of the unknown gas based on the molar mass of argon, which assumptions should be made?
Chemistry
1 answer:
vivado [14]3 years ago
8 0

Answer:

The molar mass of the unknown gas is 100.4 g/mol

Explanation:

Step 1: Data given

Molar mass of argon = 39.95 g/mol

After filling with argon the flask gained 3.221 grams

After filling with an unknown gas, the flask gained 8.107 grams

Step 2: Calculate the molar mass of the unknown gas

The gas with the higher molar mass will have the higher density.

Ar - 3.224 g; molar mass = 39.95 g/mol

X = 8.102 g; molar mass = ??

Molar mass of the unknown gas = 8.102g X *(39.95 g/mol  / 3.224 g) = 100.4 g/mol

The molar mass of the unknown gas is 100.4 g/mol

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VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

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6 0
3 years ago
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
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Answer:

0.1066 hours

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A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

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3 years ago
Bilangan oksidasi vanadium paling tinggi terdapat dalam senyawa..
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How many molecules are in 5.60 L of oxygen gas
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3 0
3 years ago
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A sample of gas occupies 10.0 L at 240°C under a pressure of
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Answer: 1090°C

Explanation: According to combined gas laws

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where P1 = initial pressure of gas = 80.0 kPa

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T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

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T2 = final temperature of gas

Substituting the values,

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T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

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3 years ago
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