According to the second order formula:
1/[At] = K t + 1/[Ao]
and when we have the K constant =0.0265 & we have t = 180 min & we have the initial concentration of A = 4.25 so by substitution:
1/[At] = 0.0265 X 180min + 1/4.25
1/[At] = 5
∴[At] = 1/5 = 0.2 m
Answer:
No
Explanation:
No, but the total mass of reactants must equal the total mass of products to be a balanced equation.
Example: Consider the following reaction ...
3H₂ + N₂ => 2NH₃ and 'amu' is atomic mass units (formula weights from periodic table)
In terms of molecules, there are 4 molecules on the left (3 molecular hydrogens (H₂) and 1 molecular nitrogen (N₂) and 2 molecules of ammonia on the right side of equation arrow. ∑reactant molecules ≠ ∑product molecules.
In terms of mass of reactants & mass of products, the 3H₂ + N₂ => 6amu + 28amu = 34amu & mass of products (2NH₃) => 2(14amu) + 6(1amu) = 34amu for sum of product masses.
∑mass reactants = ∑mass products <=> 34amu = 34amu.
The expression '∑mass reactants = ∑mass products' as applied to chemical equations is generally known as 'The Law of Mass Balance'.
Answer:
1.99grams
Explanation:
- First, we need to calculate the molar mass of the compound: Ca(HCO3)2
Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol
Hence, Ca(HCO3)2
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
Molar mass of Ca(HCO3)2 = 162g/mol
- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.
Oxygen = {16(3)}2
= 48 × 2
= 96g of Oxygen
- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.
% composition of O = 96/162 × 100
= 0.5926 × 100
= 59.26%.
- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass
0.0207 = mass/162
Mass = 162 × 0.0207
Mass = 3.353grams
However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen
Hence, in 3.353grams of Ca(HCO3)2, there will be;
0.5926 × 3.353
= 1.986
= 1.99grams.
Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.
