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3 years ago

How many molecules are in 128g of H2O?

Chemistry

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

4.28x10^24 molecules

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. From the above, we understood that 1mole of H2O also contains 6.02x10^23 molecules.

1mole of H2O = (2x1) + 16 = 2 + 16 = 18g

Now, if 18g of H2O contains 6.02x10^23 molecules,

Then 128g of H2O will contain = (128x 6.02x10^23) /18 = 4.28x10^24 molecules

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Need help ASAP! How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?

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Answer:

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Explanation:

Molarity is a measure of concentration in moles per liter.

$molarity= \frac{moles \ of \ solute}{liters \ of \ solution}$

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

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liters of solution=0.25 L
moles of solute =x

$1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}$

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

$0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L$

$0.25 \ L *1.2 \ mol \ NaNO_3/L=x$

The units of liters cancel, so we are left with the units moles of sodium nitrate.

$0.25 *1.2 \ mol \ NaNO_3=x$

$0.3 \ mol \ NaNO_3=x$

There are 0.3 moles of sodium nitrate.

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2 years ago

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(I)how many atoms are present in 7g of lithium?

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Answer :

(i) The number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

(ii) The number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

(iii) The number of moles of $F_2$ is, 1 mole

The number of moles of $CO_2$ is, 0.5 mole

The number of moles of $OH^-$ is, 1 mole

Explanation :

Part (i) :

First we have to calculate the moles of lithium.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}$

Molar mass of Li = 6.94 g/mole

$\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole$

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains $6.022\times 10^{23}$ number of atoms

So, 1.008 mole of lithium contains $1.008\times 6.022\times 10^{23}=6.07\times 10^{23}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

Part (ii) :

First we have to calculate the moles of carbon.

$\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}$

Molar mass of C = 12 g/mole

$\text{Moles of }C=\frac{24g}{12g/mol}=2mole$

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains $6.022\times 10^{23}$ number of atoms

So, 2 mole of carbon contains $2\times 6.022\times 10^{23}=1.204\times 10^{24}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

Part (iii) :

To calculate the moles of $F_2$ :

$\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}$

Molar mass of $F_2$ = 38 g/mole

$\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole$

Thus, the number of moles of $F_2$ is, 1 mole

To calculate the moles of $CO_2$ :

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mole

$\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole$

Thus, the number of moles of $CO_2$ is, 0.5 mole

To calculate the moles of $OH^-$ ions :

$\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}$

Molar mass of $OH^-$ = 17 g/mole

$\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole$

Thus, the number of moles of $OH^-$ is, 1 mole

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3 years ago