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:
K2CO3(aq) + H2O(l)
Explanation:
Balance the equation
2KOH(aq) + CO2(g) ------ K2CO3(aq) + H2O(l)
The value of current solubility product is calculated as below
K = (Ag+)( Cl-)
Ag+ = 0.01 M
Cl-= 1 x10^-5M
K is therefore = 1 x10^-5 x 0.01 = 1 x10 ^ -7 M
The K obtained is greater than Ksp
that is K> KSp
1x10^-7 > 1.7 x10 ^-10
will precipitation of AgCl form?
yes the precipitation of AgCl will be formed since K> KSP
It means what are some way we can save water I think sorry if it’s wrong.
The solution for this problem would be:
We are looking for the grams of magnesium that would have
been used in the reaction if one gram of silver were created. The computation
would be:
1 g Ag (1 mol Mg) (24.31 g/mol) / (2mol Ag)(107.87g/mol) =
0.1127 grams of Magnesium
