Gravitational Pull I’m pretty sure is the answer
Answer:
13.8072 kj
Explanation:
Given data:
Mass of water = 100.0 g
Initial temperature = 4.0 °C
Final temperature = 37.0°C
Specific heat capacity = 4.184 j/g.°C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 37.0°C - 4.0 °C
ΔT = 33.0°C
Q = 100.0 g ×4.184 j/g.°C × 33.0°C
Q = 13807.2 j
Joule to KJ:
13807.2 j × 1kj /1000 j
13.8072 kj
Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
Yes. Look up Newton’s laws of physics. That should help
