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LenaWriter [7]
3 years ago
15

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

Chemistry
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

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Studentka2010 [4]

Answer:

<em>(H30+)= 1x10^-6 M</em>

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

(H30+)= 1x10^-14/1x 10^-8

<em>(H30+)= 1x10^-6 M</em>

6 0
3 years ago
0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2Al(OH)3(s) +
Veseljchak [2.6K]

Answer:

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄ ----------------  6 moles of water

                        0.55 moles of H₂SO₄ -----------    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

5 0
3 years ago
PLEASE ANSWER
PolarNik [594]
I think its A.
if one force cannot overcome the other, the object remains stationary.
4 0
3 years ago
How many moles of ethane (C2H6) would be needed to react with 62.3 grams of oxygen gas?
Vladimir [108]
Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas

Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988

Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol

now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1

1 C2H6 + 1 O2 -> products

this means that for 1 mol of C2H6, 1 mol of O2 has to react with it

Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
5 0
3 years ago
- What is the concentration of hydrogen ions if the pH is 12?
ELEN [110]

Answer:

1 x 10^-12 mol dm^-3

Explanation:

[H+] = 10 ^ -pH

Therefore, pH 12 = 1 x 10^-12 hydrogen ion concentration

4 0
3 years ago
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