That is true b<span>ecause 1 nanometer = 1.0 × 10^-15 megameters.</span>
Answer:
1= K⁺= (Z=19) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰
2 = Zn²⁺= (Z = 30) =1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰, 3d¹⁰
Explanation:
When an atom lose or gain the electron ions are formed. There are two types of ions cation and anion.
Cation are formed when atom lose the electron.
Anion are formed when an atom gain the electron.
In given question potassium loses its valance electron and form K⁺ cation. Thus its electronic configuration will be written as,
₁₉K⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰
While the electronic configuration of potassium in neutral form is:
₁₉K = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
The atomic number of zinc is 30 and its electronic configuration is:
₃₀Zn= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², 3d¹⁰
When zinc atom loses its 2 valance electrons the electron configuration will be,
₃₀Zn²⁺= 1s² 2s² 2p⁶ 3s² 3p⁶ 4s⁰, 3d¹⁰
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.
