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:
C, 42g
Explanation:
In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).
Assuming no heat is lost to surroundings,
the energy lost from metal pellet = energy gained for water
Since E = mc∆T
(energy = mass x specific heat capacity x temperature change)
mc∆T (metal pellet) = mc∆T (water)
100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)
3958.96 = 94.14m
m = 42g
Answer:
Nitrifying Bacteria are a group of aerobic bacteria important in the nitrogen cycle as converters of soil ammonia to nitrates, compounds usable by plants. An example is nitrosomonas or nitrobacter and species in that family.
The schematic diagram is attached below, which summarises the oxidation of ammonia or free nitrogen in the soil to nitrates for the cowpea plant's utilisation.
Answer:
Ionic
Explanation:
If A does not have electron to bond, it just receives one electron from B.
It can´t be covalent because A don´t have any electrons to bond with B.
The heat required to completely melt the given substance, platinum, we just have to convert first the given mass in mole and multiply the answer to its molar heat of fusion..
Hf = mass x (1/molar mass) x molar heat of fusion
Hf = (85.5 g) x (1 mole/195.08 g) x 4.70 kcal/mol
Hf = 2.06 kcal
