Molar mass NaCl = 58 g
Mass of solute = 29 g
number of moles: mass of solute / molar mass
n = 29 / 58
n = 0.5 moles
hope this helps!
Firstly need to determine the empirical formula of the hydrocarbon. Empirical formula is the simplest whole number ratio of components of the compound. Molecular formula is the actual composition of the components in the compound.
percentage of C - 82.66%
percentage of H - (100-82.66) = 17.34 %
in 100 g of compound ;
mass of C - 82.66 g
mass of H - 17.34 g
C H
mass in 100 g 82.66 g 17.34 g
molar mass 12 g/mol 1 g/mol
number of moles 6.88 mol 17.34 mol
(mass/molar mass)
divide the number of moles by least number of moles (6.88 mol)
6.88 mol/6.88 17.34/6.88
1 2.52
multiply these by 2 to get a whole number
C - 1x 2 = 2
H - 2.52 x 2 = 5.04
round off to nearest whole number
C - 2
H - 5
ratio of C to H is 2:5
empirical formula - C₂H₅
empirical formula mass = 12 g/mol x 2 + 5 * 1 g/mol = 29 g
next have to find how many empirical units are there in the molecular unit
molecular unit mass = 58.12 g
empirical unit = 29 g
then number of empirical units = 58.12 / 29 = 2
rounded off , number of empirical units = 2
(C₂H₅) * 2 units
molecular formula = C₄H₁₀
Explanation:
Answer:
pOH = 6.73
Explanation:
First, calculate pH = -log[H+] = -log(5.4 x 10^-8) = 7.27
pOH + pH = 14, so pOH = 14 - 7.27 = 6.73
Let x be the mass of the ore in grams such that the equation that would allow us to answer this item would be,
55 = (x)(0.243)
The value of x is calculated as,
x = 55/0.243
x = 226.337 grams
Then, solve for amount in kilograms by dividing the calculated value by 1000. This methodology will give us an answer of 0.226337. Hence, the answer is 0.226.