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Natali5045456 [20]
3 years ago
13

A solid sample contains both NaOH and NaCl. 0.500 g of this solid sample was dissolved in water to make a 20.0 mL solution and t

hen this solution was titrated by 0.500 mol/L HCl solution. If 20.8 mL of HCl was used to reach the end point, what is the mass % of NaOH in the original solid sample? NaOH + HCl --> NaCl + H2O

Chemistry
1 answer:
Setler [38]3 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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How many Faradays of electricity are carried by 1.505×10^23 electronsk​
aksik [14]

Mole of electron required by 1.505 *10^{23} mole is 2.5* 10  ^{-1}

  • Faraday law expressed how the  change that is been being produced by a current at an electrode-electrolyte interface is related and  proportional to the quantity of electricity  that is been used.

  • There is one mole of electron required for 1 Faraday of electricity.

  • Avogadro constant is 6.02*10^{23}

  • Mole of electron can be calculated by dividing the number of electron by avogadro's constant.

=\frac{1.505*10^{23} }{6.02*10^{23} }

= 2.5* 10^{-1}  Faraday  of electricity

Therefore, it requires  2.5*10^{-1} Faraday of electricity for the 1.505 *  10^{23}  mole.

Learn more at: brainly.com/question/1640558?referrer=searchResults

3 0
2 years ago
If sulfur dioxide and oxygen can be made into sulfur trioxide, what is the reverse reaction? A. SO3 +O2à SO3 B. 2SO2 + O2à 2SO3
ycow [4]
REVERSE REACTION WILL BE SO2+O2
7 0
3 years ago
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8 Fe + Sg → 8 Fes<br><br> How many grams of FeS is produced from 0.3 mol Sg?
frozen [14]

Answer:

Explanation:

Molar ratio for Sg : FeS = 1:8

If there are 0.3 moles for Sg

Therefore, 0.3 × 8 =2.4 moles of FeS

Mass = Moles/ Mr

Mr of FeS = 56+32=88

So mass = 2.4/88

Mass= 0.027g

6 0
3 years ago
13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
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3 years ago
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Monica [59]

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