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3 years ago

A solid with a mass of Mgrams is melted by heating it. After a half-hour, the solid is

Chemistry

1 answer:

Lerok [7]3 years ago

7 0

Answer:

equal to M

Explanation:

The mass of the fully melted mass and the initial solid will be the same. So, the mass of the melt is equal to M.

Mass is the amount of matter contained within a substance. Since only the phase changed and the amount of matter is still the same, the mass of the molten phase and the solid phase will remain the same.

We are correct to say that in the heating process no mass was destroyed or added in melting the solid.

A simple phase change that preserved the mass only occurred.

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How much HCI exists in 275 mL of 0.83 M<br> solution of HCI?<br> Answer in units of mol.

zubka84 [21]

Answer:

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Explanation:

$M=\frac{moles}{liters} \\$

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1. Cross multiply!

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3 years ago

If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite

Elodia [21]

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of $Al_2O_3$ = 13.2 kg = 13200 g

Mass of $NaOH$ = 50.4 kg = 50400 g

Mass of $HF$ = 50.4 kg = 50400 g

Molar mass of $Al_2O_3$ = 101.9 g/mole

Molar mass of $NaOH$ = 40 g/mole

Molar mass of $HF$ = 20 g/mole

Molar mass of $Na_3AlF_6$ = 209.9 g/mole

First we have to calculate the moles of $Al_2O_3,NaOH$ and $HF$ .

$\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=\frac{13200g}{101.9g/mole}=129.54moles$

$\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{50400g}{40g/mole}=1260moles$

$\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=\frac{50400g}{20g/mole}=2520moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Al_2O_3+6NaOH+12HF\rightarrow 2Na_3AlF_6+9H_2O$

From the balanced reaction we conclude that

The mole ratio of $Al_2O_3,NaOH$ and $HF$ is, 1 : 6 : 12

And, the ratio of given moles of $Al_2O_3,NaOH$ and $HF$ is, 129.54 : 1260 : 2520

From this we conclude that, $NaOH,HF$ is an excess reagent because the given moles are greater than the required moles and $Al_2O_3$ is a limiting reagent and it limits the formation of product.