<u>Answer:</u> The actual yield of the carbon dioxide is 47.48 grams
<u>Explanation:</u>
For the given balanced equation:
To calculate the mass for given number of moles, we use the equation:
Theoretical moles of carbon dioxide = 1.30 moles
Molar mass of carbon dioxide = 44 g/mol
Putting values in above equation, we get:
To calculate the theoretical yield of carbon dioxide, we use the equation:
Theoretical yield of carbon dioxide = 57.2 g
Percentage yield of carbon dioxide = 83.0 %
Putting values in above equation, we get:
Hence, the actual yield of the carbon dioxide is 47.48 grams
None of the above the answer would be D
Answer:
n = 0.3 mol
Explanation:
Given data:
Volume of gas = 8.0 L
Temperature of gas = 45 °C (45+273 = 318 K)
Pressure of gas = 0.966 atm
Moles of gas present = ?
Ideal gas constant = R = 0.021 atm.L/mol.K
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Now we will put the values:
0.966 atm × 8 L = n × 0.0821 atm.L/mol.K × 318 K
7.728 atm.L = n × 26.12 atm.L/mol
7.728 atm.L / 26.12 atm.L/mol = n
n = 0.3 mol
The stock solution has a volume of 0.208L. When diluted, a stock solution with a concentration of 0.30 m clo2 becomes a 0.20 m stock solution with a volume of 0.050.
A three-dimensional space's occupied volume is measured. It is frequently measured numerically with SI-derived units (such the cubic metre and litre) or with other imperial units (such as the gallon, quart, cubic inch). Volume and length are related in how they are defined (cubed).
A highly concentrated stock solution, often prepared as a 10x concentrated solution, can be described as a stock solution since it can be diluted to create a working solution. In order to prepare complex solutions with numerous constituents, stock solutions can also be utilized as a component.
0.3M*v = 0.125M*0.50L
V=0.208L
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1.) Mass
2.) Can occupy space (Volume)
