Hello:
In this case, we will use the Clapeyron equation:
P = ?
n = 8 moles
T = 250 K
R = 0.082 atm.L/mol.K
V = 6 L
Therefore:
P * V = n * R * T
P * 6 = 8 * 0.082* 250
P* 6 = 164
P = 164 / 6
P = 27.33 atm
Hope that helps!
The molar mass of CO2 can be calculated as follows;
CO2 — 12 + (16x2) = 12+ 32 = 44 g
Therefore molar mass of CO2 is 44 g/mol
In 44 g of CO2 there’s 1 mol of CO2
Then 1 g of CO2 there’s 1/44 mol of CO2
Therefore in 78.3 g of CO2 there’s — 1/44 x 78.3 =1.78 mol of CO2
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004
