Answer:
212.304 grams
Explanation:
similar to your other question, use the same formula
q=mCpΔT
23617=m(4.182)(46.6-20)
23617=111.2412m
m=212.304g
Answer:
The filter bed is cleaned by occasional backwashing ;-; im sorry if this isn't a great answer but I tried
Answer:
the value of equilibrium constant for the reaction is 8.5 * 10⁷
Explanation:
Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
Given that,
We are given:
Equilibrium amount of titanium = 2.93 g
Equilibrium amount of titanium tetrachloride = 2.02 g
Equilibrium amount of chlorine gas = 1.67 g
We calculate the No of mole = mass / molar mass
mass of chlorine gas = 1.67 g
Molar mass of chlorine gas = 71 g/mol
mole of chlorine = 1.67 / 71
= 7.0L
Concentration of chlorine is = no of mole / volume
= 0.024 / 7
= 3.43 * 10⁻³M
equilibrium constant Kc =
= ![\frac{1}{[3.43 * 10^-^3]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5B3.43%20%2A%2010%5E-%5E3%5D%5E2%7D)
= 8.5 * 10⁷
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.
