Answer:
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Explanation:
Answer:
B
Explanation:
found it from a person wjwjdjjanqnqq
Answer:
975.56×10²³ molecules
Explanation:
Given data:
Number of molecules of C₂H₆ = 4.88×10²⁵
Number of molecules of CO₂ produced = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Number of moles of C₂H₆:
1 mole = 6.022×10²³ molecules
4.88×10²⁵ molecules×1mol/6.022×10²³ molecules
0.81×10² mol
81 mol
Now we will compare the moles of C₂H₆ with CO₂.
C₂H₆ : CO₂
2 : 4
81 : 4/2×81 = 162 mol
Number of molecules of CO₂:
1 mole = 6.022×10²³ molecules
162 mol ×6.022×10²³ molecules / 1 mol
975.56×10²³ molecules
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.
Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
