Q= mcΔT
1623 = 33.69g x c x (110.8 - 29.4)
1623 = 2742.366 g•°C x c
c = 0.59j/g•°C
Mass of N₂ will be formed : 50.12 g
<h3>Further explanation</h3>
Reaction
2 NH₃(g) + 3 CuO(s) → N₂(g) + 3 Cu(s) + 3 H₂O(g)
moles NH₃ = 3.58
ratio mol NH₃ : mol N₂ = 2 : 1
so mol N₂ :
mass N₂ (MW=28 g/mol) :
Answer:
4.15 M
Explanation:
In order to find the molarity of a stock solution of Z, we use the formula;
Co = 10pd/M
where;
p = percentage by mass of Z
d = density of Z
M= molar mass of Z
Substituting values;
Co= 10 * 35 * 1.4/118
C0= 4.15 M
Answer:
Explanation:
Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)
Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V
Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V
Anode reaction :
Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V
Cathode reaction :
Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V
Cell potential = Ecathode - Eanode
Ecell = .80 - ( - 2.90 )
Ecell = 3.7 V .
equilibrium constant ( K ) :
Ecell = .059 log K / n
n = 2
3.7 = .059 log K / 2
log K = 125.42
K = 2.63 x 10¹²⁵ .
Free energy change :
ΔG = - n F Ecell
= - 2 x 96500 x 3.7
= 714100 J
= 7.141 x 10⁵ J .
