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3 years ago

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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A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final

salantis [7]

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of $HNO_3$ = 13.5 M

$M_2$ = concentration of diluted solution = ?

$V_1$ = volume of $HNO_3$ = 25.0 ml = 0.0250 L

conversion used : (1 L = 1000 mL)

$V_2$ = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

$13.5M\times 0.0250L=M_2\times 0.500L$

$M_2=0.675M$

Therefore, the concentration of the diluted solution is 0.675 M

8 0

3 years ago

Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magne

Fynjy0 [20]

Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.

(a) Barium nitrate is a salt formed by the cation barium Ba²⁺ and the anion nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a salt formed by the cation potassium K⁺ and the anion chlorate ClO₃⁻. Its formula is KClO₃.

(b) The balanced equation for the decomposition of potassium chloride is:

2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)

Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)

(d) The balanced equations of metals with oxygen to form metal oxides are:

2 Mg(s) + O₂(g) ⇄ 2 MgO(s)

4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)

4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)

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3 years ago

A hydrocarbon with vapour density

Dafna1 [17]

Answer:

the answer is C4H10 I hope I helped

4 0

3 years ago

CH4 + 202 → CO2 + 2H2O How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

Step 2: Identify Conversions

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$