T = 20 % : 20 / 100 = 0.2
m1 = solute
m2 = Solvent
T = m1 / m1 + m2
0.2 = 500 g / 500 g + m2
0.2 * ( 500 + m2 ) = 500
0.2 * 500 + 0.2 m2 = 500
100 + 0.2 m2 = 500
0.2 m2 = 500 - 100
0.2 m2 = 400
m2 = 400 / 0.2
m2 = 2000 g of water
hope this helps!
Answer:
it is the app for homework not for nonsense things.
do good be good and see good and think good
The equilibrium constant for the reaction is 0.00662
Explanation:
The balanced chemical equation is :
2NO2(g)⇌2NO(g)+O2(g
At t=t 1-2x ⇔ 2x + x moles
The ideal gas law equation will be used here
PV=nRT
here n=
=
= density
P =
density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm
putting the values in reaction
0.75 = 
M = 34.61
to calculate the Kc
Kc=![\frac{ [NO] [O2]}{NO2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BNO%5D%20%5BO2%5D%7D%7BNO2%7D)
x M NO2 +
M NO+
M O2
Putting the values as molecular weight of NO2, NO,O2

34.61= 
x= 0.33
Kc= 
putting the values in the above equation
Kc = 0.00662
The volume of the balloon would decrease
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.