Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Answer:
wax i think
Explanation:
i don't understand a thing you just wrote
Answer:
<em>the </em><em>two </em><em>elements</em><em> </em><em>are </em><em>in </em><em>the</em><em> same</em><em> </em><em>period</em><em>,</em><em> with</em><em> </em><em>element </em><em>R </em><em>the </em><em>first</em><em> </em><em>element</em><em> </em><em>in </em><em>the</em><em> </em><em>period</em><em> </em><em>and </em><em>element </em><em>Q </em><em>the </em><em>last</em><em> </em><em>element</em>
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.
Resistance depends on there properties of a wire:
- length (the longer the more resistance)
- area (the less area = more resistance and vise verse more area = less resistance)
- resistivity (the more resistivity the resistance)
Hope this helps :)
