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.
Answer:
Scandium is the smallest element of the four
Explanation:
Vanadium = 50.9415 u
Titanium = 47.867 u
Scandium = 44.955912 u
Chromium = 51.9961 u
The molar mass of the gene fragment is 19182 g/mol.
What is osmotic pressure ?
Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It is also defined as the measure of the tendency of a solution to take in a pure solvent by osmosis. Potential osmotic pressure is the maximum osmotic pressure that could develop in a solution if it were separated from its pure solvent by a semipermeable membrane.
We employ the osmotic pressure equation to determine the solute's concentration, which is:
π = iMRT
Using the values in the equation above, we obtain: 19182 g/mol.
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A forest fire is considered a chemical change because of the changing of the wood by combustion, which is not a physical change.
The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
