Answer: The amount of water formed is 12 moles
Explanation: Please see the attachments below
Answer:
Ammonium chloride
Explanation:
The powder is:- Ammonium chloride
When mixed with silver nitrate, white prescipitate of silver chloride is formed as:-

When mixed with sodium hydroxide, ammonia gas is formed which has noxious order.

Ammonia gas on reaction with nickel (II) hydroxide forms deep blue colored complex as shown below:-
![Ni(OH)_2(s) + 6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}(aq) + 2OH^{-}(aq)](https://tex.z-dn.net/?f=Ni%28OH%29_2%28s%29%20%2B%206NH_3%28aq%29%5Crightarrow%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%28aq%29%20%2B%202OH%5E%7B-%7D%28aq%29)
Answer:
Option D. 17.5
Explanation:
Equiibrium is: CO + 2H₂ ⇄ CH₃OH
1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.
Initially we have 0.42 moles of CO and 0.42 moles of H₂
If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.
So in the equilibrium we may have:
0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂
Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen
Finally 0.13 moles of methanol, are found after the equilibrium reach the end.
Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²
0.13 / (0.29 . 0.16²)
Kc = 17.5
Answer: The heat required is 6.88 kJ.
Explanation:
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of ethanol = 25.0 g
= specific heat of solid ethanol= 0.97 J/gK
= specific heat of liquid ethanol = 2.31 J/gK
n = number of moles of ethanol = 
= enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole
= change in temperature
The value of change in temperature always same in Kelvin and degree Celsius.
Now put all the given values in the above expression, we get
![\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B25.0%20g%5Ctimes%200.97J%2FgK%5Ctimes%20%28-114-%28-135%29K%5D%2B0.534mole%5Ctimes%205020J%2Fmole%2B%5B25.0g%5Ctimes%202.31J%2FgK%5Ctimes%20%28-50-%28-114%29%29K%5D)
(1 KJ = 1000 J)
Therefore, the heat required is 6.88 kJ
First, we have to know that we can put [H+] instead of [H3O+]
so, according to the reaction equation:
by using ICE table:
H2O ↔ H+ + OH-
initial 0 0
change +X +X
Equ X X
when Kw = [H+] [OH-]
and when we have Kw = 2.4 x 10^-14
and when [H+] = [OH-] = X
∴ 2.4 x 10^-14 = X^2
∴ X = √(2.4 x 10^-14)
= 1.55 x 10^-7
∴[H+] = 1.55 x 10^-7