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3 years ago

Give me lil reasoning so I know your not lying for points

Chemistry

1 answer:

rjkz [21]3 years ago

5 0

1mL=0.001L
so if you have 300mL, multiply that by 0.001 to give you 0.3L

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

You're provided a 0.50 M solution of HCl. How many moles of HCl are present in 25 ml of this solution?

ddd [48]

Answer:

0.0125 moles of HCl

Explanation:

(0.50 mol/L)(0.025 L) = 0.0125 moles of HCl

4 0

3 years ago

Can someone help this is a final exam its really important

Burka [1]

Do you still need help?!?!

5 0

3 years ago

The normal boiling point of a liquid is 282 °C. At what temperature (in °C) would the vapor pressure be 0.500 atm? (∆Hvap = 28.5

NNADVOKAT [17]

For a normal boiling point of a liquid is 282 °C, the temperature is mathematically given as

T2=181.55°C\

<h3> What temperature (in °C) would the vapor pressure be 0.500 atm? </h3>

Generally, the equation for the gas is mathematically given as

ln(p1/p2)=dHvap/R(1/T2-1/T1)

Therefore

ln(1/0.26)=23500/8.214(1/T2-1/555)

T2=181.55^C

In conclusion

T2=181.55°C