Which of the following is the CORRECT path taken by sperm from site of production to site of ejection? A. epididymis, seminifero
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<u>Answer:</u> The molecular formula of the compound is
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
Or,
where,
= osmotic pressure of the solution = 2.13 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Given mass of compound = 2.04 g
Volume of solution = 175.0 mL
R = Gas constant =
T = temperature of the solution =
Putting values in above equation, we get:
- <u>Calculating the molecular formula:</u>
The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
<u>For calculating the mass of carbon:</u>
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 36.26 g of carbon dioxide, of carbon will be contained.
<u>For calculating the mass of hydrogen:</u>
In 18 g of water, 2 g of hydrogen is contained.
So, in 14.85 g of water, of hydrogen will be contained.
Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.
For Carbon =
For Hydrogen =
For Oxygen =
Converting the mole fraction into whole number by multiplying the mole fraction by '2'
Mole fraction of carbon = (1 × 2) = 2
Mole fraction of oxygen = (2.5 × 2) = 5
Mole fraction of hydrogen = (1 × 2) = 2
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 5 : 2
The empirical formula for the given compound is
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
We are given:
Mass of molecular formula = 133.9 g/mol
Mass of empirical formula = 61 g/mol
Putting values in above equation, we get:
Multiplying this valency by the subscript of every element of empirical formula, we get:
Hence, the molecular formula of the compound is