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RoseWind [281]
1 year ago
8

A student uses a solution of 1.2 molar sodium hydroxide (NaOH) to calculate the concentration of a solution of sulfuric acid (H2

SO4). She records a neutral pH after adding 20 mL of the sodium hydroxide solution to 50 mL of the sulfuric acid solution.
What is the concentration of the sulfuric acid solution?
A.
0.24 M
B.
0.12 M
C.
2.1 M
D.
1.0 M
E.
0.48 M
Chemistry
1 answer:
Mila [183]1 year ago
5 0

From the calculations, the concentration of the acid is 0.24 M.

<h3>What is neutralization?</h3>

The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.

We have to use the formula;

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O

CA = ?

CB = 1.2 M

VA =  50 mL

VB = 20 mL

NA = 1

NB = 2

CA = CBVBNA/VANB

CA = 1.2 M * 20 mL * 1/ 50 mL * 2

CA = 0.24 M

Learn more about neutralization:brainly.com/question/27891712

#SPJ1

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5 0
3 years ago
CS2 (s) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)
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Answer:

2.067 L ≅ 2.07 L.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>

It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

  • At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u><em>using cross multiplication:</em></u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

  • To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

<u><em>Using cross multiplication:</em></u>

3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

  • Again, using cross multiplication:

1.0 mol of SO₂ represents → 22.4 L, at STP.

0.09227 mol of SO₂ represents → ??? L.

∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.

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3 years ago
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2 years ago
What forces hold network solids together?
ZanzabumX [31]

Answer:

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8 0
3 years ago
How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

4 0
3 years ago
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