Answer:
301.8 g
Explanation:
We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.
m(solution) = m(solute) + m(solvent)
m(solution) = 200.4 g + 101.42 g
m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)
Answer:
2K + Br2 --> 2KBr is the answer
They shots sides stoned found six end
Answer:
0.302L
Explanation:
<em>...97.1mL of 1.21m M aqueous magnesium fluoride solution</em>
<em />
In this problem the chemist is disolving a solution from 1.21mM = 1.21x10⁻³M, to 389μM = 389x10⁻⁶M. That means the solution must be diluted:
1.21x10⁻³M / 389x10⁻⁶M = 3.11 times
As the initial volume of the original concentration is 97.1mL, the final volume must be:
97.1mL * 3.11 = 302.0mL =
0.302L
<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
