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ikadub [295]
2 years ago
9

Can someone help me please?

Chemistry
1 answer:
Deffense [45]2 years ago
4 0
It doesn't exactly tell you the story...
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4 0
3 years ago
How many digits to the right of the decimal point should be used to report the result?
aalyn [17]

Answer:

In engineering and science the common stand is two places.

For example if you get a calculation of 4.567 round up and give the result of 4.57

4 0
3 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
2 years ago
2KI + Pb(NO3)2 → 2KNO3 + PbI2 Determine how many moles of KNO3 are created if 0.03 moles of KI are completely consumed.
ElenaW [278]
2:2 so the same proportion 
0,03

6 0
3 years ago
Read 2 more answers
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