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Kruka [31]
2 years ago
10

One method used commercially to peel potatoes is to soak them in a solution of NaOH for a short time and then remove them from t

he NaOH. The peel will have separated from the potato and so it can then easily be sprayed off to yield beautifully peeled potatoes! The NaOH solution however, must be analyzed periodically to make sure its concentration is adequate for the reaction In one such analysis, 45.7 mL of 0.500 M H2SO4 is required to react completely with a 20.0 mL sample of the NaOH solution. The products of the reaction are Na2SO4 and water. What is the concentration of the starting NaOH solution?a. 1.14 M b. 2.28 M c. 0.438 M d. 0.500 M e. 0.571M
Chemistry
1 answer:
sergij07 [2.7K]2 years ago
7 0

Answer:

b. 2.28 M

Explanation:

The reaction of neutralization of NaOH with H2SO4 is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

<em>Where 2 moles of NaOH react per mole of H2SO4</em>

<em />

To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:

<em>Moles H2SO4:</em>

45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4

<em>Moles NaOH:</em>

0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH

<em>Molarity NaOH:</em>

0.0457moles NaOH / 0.020L =

2.28M

Right option:

<h3>b. 2.28 M</h3>
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3 years ago
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A compound is composed of 58.8% c, 9.8% h, and 31.4% o, and the molar mass is 102 g/mol. what is the molecular formula for this
Ivahew [28]

The solution would be like this for this specific problem:

 

<span>Moles of carbon = 58.8 / 12 = 4.9 </span><span>
<span>Moles of hydrogen = 9.8 / 1 = 9.8 </span>
<span>Moles of oxugen = 31.4 / 16 m= 1.96 </span>
<span>Ratio 4.9 / 1.96 = 2.5 9.8 / 1.96 = 5.0 1.96 / 1.96 = 1 </span></span>

Simplest formula = C5H10<span>


</span><span>I hope this helps and if you have any further questions, please don’t hesitate to ask again.</span>
8 0
3 years ago
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Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0
3 years ago
You have a 25.2 L sample of gas at 1.25 atm and 25.0 degrees Celsius. How many moles are present in this gas. For your answer, p
Elenna [48]

Answer:

  • <u>1.29 mol</u>

Explanation:

This is a direct application of the equation for ideal gases.

  • PV=nRT

Where:

  • P = pressure = 1.25 atm
  • V = volume = 25.2 liter
  • R = Universal constant of gases = 0.08206 atm-liter/K-mol
  • T = absolute temperature = 25.0ºC = 25 + 273.15 K = 298.15 K
  • n = number of moles

Solving for n:

  • n=\frac{PV}{RT}

Substituting:

n=\frac{1.25atm\times 25.2liter}{0.08206atm-liter/K-mol\times298.15K }\\\\n=1.29mol

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3 years ago
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Daniel [21]
3 because I just did it
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3 years ago
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