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3 years ago

What is the pOH of a 5.6 x 10-5 M solution of cesium hydroxide (CsOH)?

Chemistry

1 answer:

Rudiy273 years ago

7 0

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

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Please help will give brainliest and brainly points tyty <3

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Convection

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What would you observe when zinc is added to solutions of iron II sulphate? write the chemical reaction that takes place.

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When zinc is added to the solution of iron sulphate, the colour of iron sulphate solution changes. It is because zinc is more reactive than iron, it displaces iron from its solution of iron sulphate and a grey precipitate of iron and a colourless solution of zinc sulphate is formed.

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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 $Cl_2=0.280 M -0.228 M=0.052M$ .

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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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