Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:
![Na_2S_2O_3 => 2Na^+_{(aq)} + S_2O^2^-_3_{(aq)}](https://tex.z-dn.net/?f=Na_2S_2O_3%20%20%3D%3E%202Na%5E%2B_%7B%28aq%29%7D%20%2B%20S_2O%5E2%5E-_3_%7B%28aq%29%7D)
The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M
Answer:
D.
Explanation:
That would be the last option.
Repeated similar experimental outcomes are confirmation of the theory.
The correct option is C.
For all liquids, at boiling point the temperature remain constant while the volume of the liquid will decrease.
The temperature applied at boiling point is used to break the inter molecular forces that are holding the particles of the liquid together.
The volume of the liquid is decreasing because the particles are changing to gas and escaping to the atmosphere as gas.
Answer:
a geyser is a spring and it is a discharge of water that ejects turbulently and is accompanied by steam
Answer:
(a) The concentration of
after 5.00 min is 0.9672 mol/L
(b) The fraction of
decomposed after 5.00 min is 0.568
The problem can be solved by using first order integrated reaction.
Explanation:
(a)
Rate constant of reaction = K =
![s^{-1}](https://tex.z-dn.net/?f=s%5E%7B-1%7D)
Initial concentration of
= 2.24 mol/L
Assuming final concentration of
to be x mol/L
time (t) = 5.00 min
The first order integrated equation is shown below
![\textrm{t} = \displaystyle \frac{2.303}{K}\textrm{log}\frac{\textrm{Initial concentration of } N_{2}O_{5}}{\textrm{Final concentration of } N_{2}O_{5}} \\\left ( 5.00\times 60 \right )\textrm{ sec} = \displaystyle \frac{2.303}{2.8\times 10^{-3}s^{-1}}\textrm{log}\frac{2.24 \textrm{ mol/L}}{x} \\x = 0.9672 \textrm{ mol/L}](https://tex.z-dn.net/?f=%5Ctextrm%7Bt%7D%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B2.303%7D%7BK%7D%5Ctextrm%7Blog%7D%5Cfrac%7B%5Ctextrm%7BInitial%20concentration%20of%20%7D%20N_%7B2%7DO_%7B5%7D%7D%7B%5Ctextrm%7BFinal%20concentration%20of%20%7D%20N_%7B2%7DO_%7B5%7D%7D%20%5C%5C%5Cleft%20%28%205.00%5Ctimes%2060%20%5Cright%20%29%5Ctextrm%7B%20sec%7D%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B2.303%7D%7B2.8%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7D%5Ctextrm%7Blog%7D%5Cfrac%7B2.24%20%5Ctextrm%7B%20mol%2FL%7D%7D%7Bx%7D%20%5C%5Cx%20%3D%200.9672%20%5Ctextrm%7B%20mol%2FL%7D)
Concentration of
decomposed = 0.9672 mol/L
(b)
Concentration of
decomposed = ![\left ( 2.24-0.9672 \right )\textrm{ mol/L} = 1.273 \textrm{ mol/L}](https://tex.z-dn.net/?f=%5Cleft%20%28%202.24-0.9672%20%5Cright%20%29%5Ctextrm%7B%20mol%2FL%7D%20%3D%201.273%20%5Ctextrm%7B%20mol%2FL%7D)
Fraction of
decomposed = ![\displaystyle \frac{1.273 \textrm{ mol/L}}{2.24 \textrm{ mol/L}} = 0.568](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1.273%20%5Ctextrm%7B%20mol%2FL%7D%7D%7B2.24%20%5Ctextrm%7B%20mol%2FL%7D%7D%20%3D%200.568)