All categories

3 years ago

A 100.0-g sample of sodium hydroxide contains sodium, oxygen, and hydrogen. The

Chemistry

1 answer:

enyata [817]3 years ago

3 0

Answer:

57.47 grams of sodium is in the sample.

Explanation:

Given that a 100.0-g sample of sodium hydroxide contains sodium, oxygen, and hydrogen, and the sample contains 2.53 g of hydrogen and 40.0 g of oxygen, to determine what mass of sodium is in the sample the following calculation must be performed:

100 - (40 + 2.53) = X

100 - 42.53 = X

57.47 = X

Therefore, 57.47 grams of sodium is in the sample.

You might be interested in

Draw models to represent the formation of the<br> positive calcium ion and the negative bromide ion.

Andrej [43]

What is love baby dont hurt me dont hurt me

4 0

2 years ago

A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation o

statuscvo [17]

Answer : The molarity of $Pb(NO_3)_2$ solution is, 0.352 M

Explanation :

First we have to calculate the moles of $PbCl_2$

$\text{Moles of }PbCl_2=\frac{\text{Given mass }PbCl_2}{\text{Molar mass }PbCl_2}$

Molar mass of $PbCl_2$ = 278.1 g/mol

$\text{Moles of }PbCl_2=\frac{19.58g}{278.1g/mol}=0.07041mol$

Now we have to calculate the moles of $CaCl_2$

The balanced chemical equation is:

$Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 1 mole of $PbCl_2$ produced from 1 mole of $Pb(NO_3)_2$

So, 0.07041 mole of $PbCl_2$ produced from 0.07041 mole of $Pb(NO_3)_2$

Now we have to calculate the molarity of $Pb(NO_3)_2$

$\text{Molarity of }Pb(NO_3)_2=\frac{\text{Moles of }Pb(NO_3)_2}{\text{Volume of solution in (L)}}$

$\text{Molarity of }Pb(NO_3)_2=\frac{0.07041mol}{0.200L}=0.352M$

Therefore, the molarity of $Pb(NO_3)_2$ solution is, 0.352 M

4 0

3 years ago

PLEASE HELPPPP what characteristics of the bond you choose are persent in H2O.

Andrej [43]

Water molecules forming hydrogen bonds with one another. The partial negative charge on the O of one molecule can form a hydrogen bond with the partial positive charge on the hydrogens of other molecules. Water molecules are also attracted to other polar molecules and to ions.

6 0

3 years ago

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

5 0

2 years ago

Why Group 18 elements on the periodic table rarely form compounds

lawyer [7]

Most elements on group 18 are the Noble Gases. They already have a complete last level with 8 electrones. Actually they can form compounds but only on the lab and they will not even last half a second.

4 0

3 years ago