Answer:
Formula mass = 58.09 g
Explanation:
Formula mass of a given molecule is defined as the sum of atomic masses of elements forming the empirical formula of that molecule.
In order to calculate the formula mass of propanol, following data is required;
Empirical Formula of Propanol:
The empirical formula of Propanol is C₃H₆O (also the molecular formula) as it is the lowest possible atomic ratio of the given elements.
Atomic Masses of Elements:
Carbon = 12.01 g/mol
Hydrogen = 1.01 g/mol
Oxygen = 16.00 g/mol
Hence,
Formula mass = (At. mass of C)₃ + (At. mass of H)₆ + (At. mass of O)
Formula mass = (12.01)₃ + (1.01)₆ + (16.00)
Formula mass = 36.03 + 6.06 + 16.00
Formula mass = 58.09 g
Answer:
The correct answer is 16 gram per mole.
Explanation:
Let A be the gas helium, and B be the unknown gas. It is clearly mentioned in the question that the effusion rate of helium gas is two times more than that of gas B. The molar mass of helium is 4.0 gram per mole. To solve the problem, Grahm's law is used, that is,
Rate of effusion A/rate of effusion B = √ (Molar mass B/Molar mass of A
2.0 = √Molar mass of B/4.0 gram per mole.
Now squaring both the sides we get,
4.0 = Molar mass of B / 4.0
The molar mass of B = 16 gram per mole.
439.3 g CO2
Explanation:
First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:
3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)
= 9.981 mol CO2
Use the molar mass of CO2 to determine the # of grams of CO2:
9.981 mol CO2 x (44.01 g CO2/1 mol CO2)
= 439.3 g CO2
Answer:
The correct answer is B) it helps to ensure the result are consistent and repeatable.
Explanation:
Scientist generally repeat an experiment if he or she did not make a mistake in the first one to compare the results of two experiment, if there is no difference in the result or values obtainted from the observation of two experiment. he or she become sure that experiment was done in a right way because if there is some error made when experiment was carried out then the result of two same experiment would be different.
