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3 years ago

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A sample of oxygen gas occupies a volume of 160 liters at 364 K. What will be the volume of the gas when the temperature drops t

o 273 K?
210 L

120 L

62 L

470 L

1 answer:

Ivahew [28]3 years ago

8 0

Answer:

62L

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tysm.

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hope it help

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1 example of o neutral

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NaCl and CO are two neutral compounds as well as these are popular and these Neutral substances are those which have a pH of 7 even they are neither acidic nor basic. The substances which are neither acidic, nor basic and these are called neutral substance. For example distilled water, sugar solution etc. 3.9. 17 votes.

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The density of copper is 8.9 g/cm3.If you have 90.0 grams of copper ,how much volume is your copper occupying

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Following reactions<br>FeSO4+KOH<br>

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3 years ago

The equilibrium constant, K., for the following reaction is 83.3 at 500 K. PC13(g) + Cl2(g) = PC13(E) Calculate the equilibrium

lord [1]

Answer:

The equilibrium concentration of $PCl_5=0.228 M$ .

The equilibrium concentration of $PCl_3=0.280 M -0.228 M=0.052M$ .

The equilibrium concentration of $Cl_2=0.280 M -0.228 M=0.052M$ .

Explanation:

Answer:

The equilibrium concentration of HCl is 0.01707 M.

Explanation:

Equilibrium constant of the reaction = $K_c=83.3$

Moles of $PCl_3 = 0.280 mol$

Concentration of $[PCl_3]=\frac{0.280 mol}{1.00 L}=0.280 M$

Moles of $Cl_ = 0.280 mol$

Concentration of $[Cl_2]=\frac{0.280 mol}{1.00 L}=0.280M$

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 0.280 0.280 0

At eq'm: (0.280-x) (0.280-x) x

We are given:

$[PCl_3]_{eq}=(0.280-x)$

$[Cl_2]_{eq}=(0.280-x)$

$[PCl_5]_{eq}=x$

Calculating for 'x'. we get:

The expression of $K_{c}$ for above reaction follows:

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

Putting values in above equation, we get:

$83.3=\frac{x}{(0.280-x)\times (0.280-x)}$

On solving this quadratic equation we get:

x = 0.228, 0.344

0.228 M < 0.280 M< 0.344 M

x = 0.228 M

The equilibrium concentration of $PCl_5=0.228 M$ .

The equilibrium concentration of $PCl_3=0.280 M -0.228 M=0.052M$ .

The equilibrium concentration of $Cl_2=0.280 M -0.228 M=0.052M$ .

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3 years ago