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2 years ago

14

Help. chemistry...........

2 answers:

nasty-shy [4]2 years ago

8 0

<span>The </span>equilibrium<span> will </span>shift<span> to favor the side of the reaction that involves fewer moles of gas.
Its C

</span>

balandron [24]2 years ago

3 0

C. The equilibrium would shift in the direction of the change

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What is the volume of a 0.1 M HCl solution containing 1.46 moles of HCI?

Natasha_Volkova [10]

N = cv
1.46 = 0.1 x v
v = 1.46/0.1
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2 years ago

How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?

Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

$pH= -log[H]\\pH= -log (\frac{kw}{[OH]})$

$8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}$

$[OH]=1.69E-6$

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

$Cu(OH)_2 -> Cu^{2+} + 2OH^-$

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

$Kps= [Cu^{2+}] [OH]^2$

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

$Kps= s*(2s+1.69E-6)^2$

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is $s=4.96E-8 mol/L$

Now, let us calculate the moles in 1 L:

$moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles$

7 0

3 years ago

Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise

USPshnik [31]

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

2 N2O5(g) --> 4 NO2 + O2

Initial: 0.110 - -

change: -2x +4x +x

Final: 0.110 - 2x +4x +x

But final = 0.150atm;

0.110 - 2x + 4x + x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k = rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3) t

upon solving, t = 37.1 s

3 0

2 years ago