Answer:
300 kg
Explanation:
The number of people in the city, p = 400,000
The number of capsules of oseltamivir each person consumes per day, n = 2 capsules
The number of days each person consumes the oseltamivir capsules, t = 5 days
The mass of oseltamivir in each capsule, m = 75 mg
The mass of oseltamivir needed to treat all the people in the city, <em>M</em>, is given as follows;
M = n·t·m·p
∴ M (in milligrams) = (2 × 5 × 75 × 400,000) mg = 300,000,000 mg
1,000,000 mg = 1 kg
∴ 300,000,000 mg = 300 kg
The mass of oseltamivir needed to treat all the people in the city, <em>M</em> = 300 kg
Answer:
There's no picture, so I can't help witht his, apologies!
Explanation:
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.
Answer:
Krypton has 4 orbital shells
<u>Extra Information:</u>
<em>Krypton has 36 electrons which are arranged in these 4 orbit shells, </em>
<em>The order in which electrons are arranged in these shells is 2,8,18,8</em>
