Answer:
- % Cobalt (II) Nitrate = 30.62%
Explanation:
To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:
- Mass Water ⇒ 0.350 kg Water = 350 g water
- Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol = 91.8 g
- Mass cobalt (II) nitrate ⇒ 195.0 g
Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate
- Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g
To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:
- % Water ⇒ 350/636.8 * 100% = 54.96%
- % Ammonia ⇒ 91.8/636.8 * 100% = 0.14%
- % Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%
# of atoms per mol = Avogadro’s # (6.022 x 10^23)
Number of mols = mass of substance / molar mass
73 g / 40.08 g = 1.8 mols of Ca in 73 grams
1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
28%
Explanation:
mass of solute(KBr) = 3.73g
mass of solvent(H2O) = 131g
mass of solution = mass of solute + mass of solvent
= 3.73 + 131
= 134.73g
The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
