Answer:
To calculate the percent composition, we need to know the masses of C, H, and O in a known mass of C9H8O4. It is convenient to consider 1 mol of C9H8O4 and use its molar mass (180.159 g/mole, determined from the chemical formula) to calculate the percentages of each of its elements:
%
C
9
mol C
×
molar mass C
molar mass
C
9
H
18
O
4
×
100
=
9
×
12.01
g/mol
180.159
g/mol
×
100
=
108.09
g/mol
180.159
g/mol
×
100
%
C
60.00
%
C
Answer:
[NH3] = 0.270M
[NH4Cl] = 0.327M
Explanation:
The HNO3 will react with the weak base, NH3, as follows:
HNO₃ + NH₃ → NH₄⁺ + NO₃⁻
Initial moles of each specie of the buffer:
NH3 = NH4⁺ 0.210L * (0.300mol/L) = 0.063moles
The moles added of HNO3 = Additional moles of NH4Cl and the moles substracted of NH3:
0.001L * (6mol / L) = 0.006 moles.
After the addition:
Moles NH3 = 0.063mol - 0.006mol = 0.057moles
Moles NH4Cl = 0.063mol + 0.006mol = 0.0069moles
And their concentrations are:
[NH3] = 0.057moles / 0.211L = 0.270M
[NH4Cl] = 0.069moles / 0.211L = 0.327M
Soil pollution with composting, Air pollution with gas adsorption and water pollution with settling
Answer:
Option D is correct
Explanation:
The individual mass of reactants in the reaction are
Calcium Oxide (CaO) = 
amu
Carbon dioxide (CO2) 
amu
Since this reaction is a simple addition reaction in which no mass of the reactants is wasted. Also, no energy is released accompanied by conversion of mass into energy.
Hence, total mass of reactants would be equal to the mass of the product.
Thus, mass of calcium carbonate i.e CaCO3 is
amu
Hence, option D is correct
Componys and rocks hope that helped
