Answer:
4.16 * 10²⁴ particles
Explanation:
1. We convert from grams to moles
4 g NaCl / 58 g of NaCl = 0.0689655172 moles
-> 0.0690 moles
2. Now we will convert from moles to particles by multiplying by avogadro's number..
0.0690 * 6.022 *10²³ = 4.15518 * 10²⁴ particles or 4.16 * 10²⁴ particles
A) 2H₂(g) + O₂(g) → 2H₂O(l) + 285.83 kJ
Exothermic
B) 2Mg + O₂ → 2MgO + 1200kJ
Exothermic
Wdym mole like the animal or something else ?
B ............ is the answer,
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.
