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jeyben [28]
1 year ago
11

If 57.3 l of 0.497 m koh is required to completely neutralize 39.5 l of a CH3COOH solution. What is the molarity of the acetic a

cid solution?
Chemistry
1 answer:
bearhunter [10]1 year ago
6 0

The molarity of the solution will be 0.72 m.

The majority of reactions take place in solutions, making it crucial to comprehend how the substance's concentration is expressed in a solution when it is present. The number of chemicals in a solution can be stated in a variety of ways, including.

The symbol for it is M, and it serves as one of the most often used concentration units. Its definition states how many moles of solute there are in a liter of solution.

Given data:

V_{1} =57.3 L\\V_{2} = 39.5 L\\M_{1} = 0.497 m\\\\M_{2} = ?

Molarity can be determined by the formula:

M_{1} V_{1} = M_{2} V_{2}

where, M is molarity and V is volume.

Put the value of given data in above equation.

57.3 × 0.497 m = M × 39.5 L

M = 0.72 m

Therefore, the molarity of the solution will be 0.72 m

To know more about molarity

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The best answer would be A.) stay in the car.

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6 0
3 years ago
How many g of MgCO3(s) are needed to make 1.2 L of 1.5 M MgCl2(aq) solution?
maw [93]
Molar mass of MgCO3 is 84.313 g/mol
You can calculate this from data on the periodic table:
Molar mass Mg = 24.305g/mol
molar mass C = 12.011g/mol
molar mass O = 15.999g/mol mass 3 mol = 47.997g
Total = 84.313g/mol

Mass to be used in 1.2L of 1.5M solution = 84.313g * 1.2L * 1.5mol /L = 151.763g
I have not taken significant figures into account
The balanced equation you provide is not necessary in this calculation
4 0
3 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
WRITING AND BALANCING EQUATIONS
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Answer:

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Explanation:

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A Y particle is a what ​
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Answer:

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