Answer:
The correct answer is 6.65 mg.
Explanation:
Based on the given information, the number of days given is 45.6 days. The original mass of sodium phosphate give is 175 mg. The half-life of phosphorus-32 is 14.3 days, therefore, the n or the number of half life will be,
n = 45.6/14.3 = 3.19
So, after 45.6 days, the mass of sodium phosphate sample left will be,
= Original mass × 1/2ⁿ
= 175 mg × 1/2 ^3.19
= 175 mg × 0.1096
= 34.3 mg of sodium phosphate left after 45.6 days
The molecular mass of Na₃³²PO₄ is 3 (23) + 32 + 4 (16) amu = 165 amu
Therefore, 165 grams of sodium phosphate comprise 32 grams of phosphorus.
So, 34.3 mg or 0.343 gram of sodium phosphate will contain,
= 0.343 × 32/165
= 6.65 mg of P³².
Answer:
233 g.
Explanation:
The solubility of silver nitrate at 20.0°C is 222 g per 100 g of water.
The solubility of silver nitrate at 50.0°C is 455 g per 100 g of water.
<em>∴ We need to add (455 g - 222 g = 233 g) of AgNO₃ to obtain a saturated solution at 50.0°C.</em>
Copper reacts with Nitrogen to produce Copper Nitride
6 Cu + N₂ → 2 Cu₃N
At S.T.P
22.4 L (1mole) of N₂ reacts with = 381.24 g of Cu to produce Cu₃N
So,
33.9 L of N₂ will react with = X g of Cu
Solving for X,
X = (33.9 L × 381.24 g) ÷ 22.4 L
X = 576.96 g of Cu
As,
Moles = Mass / M.mass
Moles = 576.96 g / 63.54 g/mol
Moles = 9.08 moles of Cu
Result:
So, 33.9 L of N₂ will react with 576.96 g (9.08 moles) of Cu.
<span>The chlorine isotope with 18 neutrons has an abundance of 0.7577 and a mass number of 35 amu. To calculate the averageatomic mass, multiply the fraction by the mass number for eachisotope, then add them together.</span>
Alkaline earth metals
group 2 (IIA)