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.
What are you asking???? If the formula had no atoms of oxygen then......
Answer:
I think the answer is boiling
Answer:
............................................................................
Explanation:
Answer:
mass of HNO₃ = 0.378 g
Explanation:
Normality = Molarity * number of equivalents
Molarity = Normality/number of equivalents
normality of HNO₃ = 0.30 N, Volume = 20 mL
HNO₃ ionizes in the following way:
HNO₃(aq) ----> H⁺ + NO₃⁻
Therefore, number of equivalents for HNO₃ is 1
molarity of HNO₃ = 0.30/1 =0.30 mol/dm³
Using the formula, molarity = number of moles/volume in liters
number of moles = molarity * volume
Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL
number of moles = 0.006 moles
From the formula, mass = number of moles * molar mass
molar mass of HNO₃ = 63.0 g/mol
mass = 0.006 * 63
mass of HNO₃ = 0.378 g
