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2 years ago

Q11 : why b is correct

Chemistry

1 answer:

german2 years ago

4 0

B is correct because it doesn't depend on the movement just how the partials move at boiling point, ie it will expand but not gain motion

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A. Density of Liquids

morpeh [17]

Answer:

Mass of liquid: 20.421g

Density= 1.0109405940594 g/mL

Explanation:

Mass of liquid

To find mass of liquid you take the mass of beaker + liquid (171.223g) and subtract it from the Mass of beaker (beaker without the water). The difference is the answer.

171.223g - 150.802g = 20.421g

Density

To find density you use the formula Mass/Volume. Take the Volume given, and the mass of the liquid you just found.

20.421mL/20.421g = 1.0109405940594 g/mL

5 0

3 years ago

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2(g)+C(s)↽−−⇀CS2(g)????c=9.40 at 900 K Ho

Maru [420]

Answer : The mass of $CS_2$ is, 555.028 grams

Explanation :

First er have to calculate the concentration of $S_2$ .

$\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L$

Now we have to calculate the concentration of $CS_2$ .

The given balanced chemical reaction is,

$S_2(g)+C(s)\rightleftharpoons CS_2(g)$

Initial conc. 1.51 0 0

At eqm. conc. (1.51-x) (x) (x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CS_2]}{[S_2]}$

Now put all the given values in this expression, we get :

$9.40=\frac{x}{(1.51-x)}$

By solving the term 'x', we get :

x = 1.365 M

Concentration of $CS_2$ = x M = 1.365 M

Now we have to calculate the moles of $CS_2$ .

$\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole$

Now we have to calculate the mass of $CS_2$ .

Molar mass of $CS_2$ = 76 g/mole

$\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g$

Therefore, the mass of $CS_2$ is, 555.028 grams

3 0

4 years ago

The diagram shows one period of the Periodic Table. Li ,Be ,B , C,N,O,F,Ne .Which two elements form acidic oxides?

Rudik [331]

Answer:

A -carbon and lithium.

Explanation:

6 0

2 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.

7 0

4 years ago

Hunter is copying an angle. His work so far follows. Explain the importance of his next step. Which is drawing A-line through A

forsale [732]

Answer:

its c ma man .

This is the other ray that will make up the angle ∠AYZ and will complete the constructioExplanation:

if your taking the quick check these were my awnsers

1.c

2.b

3.d

4.c

5 .A

for the copying an angle quick check only

6 0

3 years ago