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3 years ago

Describe how fuel oil can be changed into gasoline?

1 answer:

4 0

By fractional distalation which is heating something up so it boils and evaporate then the gas condenses and runs down the tube for petrol.

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50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?

d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

$50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3$

Then you have to convert mL to L.

$250mL*\frac{1L}{1000mL} = 0.250L$

Now divide the mol by the L.

$\frac{0.588mol NaNO3}{0.250L} = 2.352 M$

Round to the smallest number of significant figures = 2.35M

7 0

3 years ago

A jeweler guarantees that a piece of jewelry is at least 95% gold, by mass. You consider buying a piece of gold jewelry that wei

ELEN [110]

The density of the sample is:
Density = mass / volume
Density = 9.85 / 0.675
Density = 14.6 g/cm³

If the sample has 95% gold, and 5% silver, its density should be:
0.95 x 19.3 + 0.05 x 10.5
Theoretical density = 18.9 g/cm³

The difference in theoretical and actual densities is very large, making it likely that the jeweler was not telling the truth.

8 0

3 years ago

Please write the process (how soap is made?.......................

Mamont248 [21]

Soap is created by mixing fats and oils with a base as opposed to detergent which is created by combining chemical compounds in a mixer.

4 0

3 years ago

Read 2 more answers

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$