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3 years ago

Please help me. My chemistry teacher is not teaching us doing this epidemic

Chemistry

1 answer:

anygoal [31]3 years ago

6 0

Answer:

The answer to your question is below

Explanation:

1.-

Data

[HCl] = 0.3 N

Volume a = 6 ml

[NaOH] = ?

Volume b = 12 ml

Formula

[HCl] x Volume a = [NaOH] x Volume b

-Solve for [NaOH]

[NaOH] = [HCl] x Volume a / Volume b

-Substitution

[NaOH] = 0.3 x 6 / 12

-Simplification

[NaOH] = 0.15 N

2.-

Data

[H₂SO₄] = 7 N

Volume a = 12 ml

[KOH] = 3 N

Volume b = ?

Formula

[H₂SO₄] x Volume a = [KOH] x Volume b

-Solve for Volume b

Volume b = [H₂SO₄] x Volume a / [KOH]

-Substitution

Volume b = 7 x 12 / 3

-Simplification

Volume b = 28 ml

You might be interested in

How many moles of HCl would react with 37.1 mL of 0.138 M Sr(OH)2

Lena [83]

Answer is: 0.102 moles of HCl would react.

Balanced chemical reaction:

2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).

V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.

V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.

c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.

n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).

n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.

n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.

From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.

n(HCl) = 2 · n(Sr(OH)₂).

n(HCl) = 2 · 0.0051 mol.

n(HCl) = 0.0102 mol; amount of the hydrochloric acid.

5 0

3 years ago

A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.

Lilit [14]

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

$\frac{V1}{T1} = \frac{V2}{T2}$

Solve for V2

V2 = $\frac{V1T2}{T1}$

3.- Substitution

V2 = $\frac{(42)(233)}{293}$

4.- Simplification

V2 = $\frac{9786}{293}$

5.- Result

V2 = 33.4ml

3 0

2 years ago

How many grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline?

egoroff_w [7]

Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g

Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,

2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0

Molar mass of octane = 114.23g/mol

Molar mass of Oxygen = 32g/mol

According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25

2 mole of octane needs 25 mole of oxygen

1 mole of octane needs 12.5 moleof oxygen

114.23g of octane needs 400g of oxygen

13g of octane needs 45.5g of oxygen

Mass of oxygen needed =45.5g

Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.

Learn more about Octane here, brainly.com/question/21268869

#SPJ4

5 0

1 year ago

Which electron dot diagram represents H2?

Marizza181 [45]

Answer:

H:H

Explanation:

7 0

2 years ago

Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i

g100num [7]

It's simple, just follow my steps.

1º - in 1 L we have $0.0100~g$ of $BaCO_3$

2º - let's find the number of moles.

$\eta=\frac{m}{MM}$

$\eta=\frac{0.0100}{197.3}$

$\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}$

3º - The concentration will be

$C=5.07\times10^{-5}~mol/L$

But we have this reaction

$BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}$

This concentration will be the concentration of $Ba^{2+}~~and~~CO_3^{2-}$

$K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}$

considering $[BaCO_3]=1~mol/L$

$K_{sp}=[Ba^{2+}][CO_3^{2-}]$

and

$[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L$

We can replace it

$K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})$

$K_{sp}\approx25.70\times10^{-10}$

Therefore the $K_{sp}$ is:

$\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}$

4 0

3 years ago

Read 2 more answers