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Ulleksa [173]
3 years ago
15

What are 3 Properties of Convalent electrons

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
4 0
Do not ionize in solutions
Poor conductors of electricity/heat
Low melting/boiling points
gases or liquids at room temperature
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Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) H 2O (l) Ca(OH) 2 (s) In a parti
Nimfa-mama [501]
Find the moles of CaO

divide mass (2.0g) by the RFM which is 56 (Ca is 40 add that to O which is 16 making 56) this gives 0.0356 moles.

Find the theoretical mass by multiplying the moles of CaO (which is 0.0356 as there are no balancing number making the ratio 1:1) by the RFM of Ca(OH)2 which is 74 (40+16+16+1+1)

74 (Ca(OH)2 RFM) x 0.0357 (CaO moles) = 2.6g which is the theoretical mass of Ca(OH)2

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8 0
4 years ago
A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t
Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess AgNO_3 then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = \frac{x}{58.5}moles

Moles of chloride ions in KCl = \frac{(0.9440-x)}{74.5}moles

The total moles of chloride ions = \frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = \frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles

Now we have to determine the value of 'x'.

Moles of AgCl = \frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles

0.0133 mole = \frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles

By solving the term, we get the value of 'x'.

x=0.171g

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%

\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

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