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stellarik [79]
3 years ago
11

When two clear, colourless liquids are mixed together, a yellow precipitate is formed. This is an example of a

Chemistry
1 answer:
Flura [38]3 years ago
3 0

Answer:

Combining the two clear colorless liquids is a chemical change.

Explanation:

Because a different solid substance is formed. Tell students that a precipitate is an insoluble solid that forms when two solutions are combined and react chemically. Insoluble means that the solid will not dissolve.

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Answer: arrow 4

Explanation:

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What is graphite and what do you use it for?
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for the given reaction, what volume of o2 would be required to react with 4.8 l of co , measured at the same temperature and pre
SVEN [57.7K]

We can see that 2 moles of The no react with 1 mole of O2 using this equation. 4.8 L NO x 1 L O2 / 2 L NO = 2.4 L of O2 are needed at constant pressure and temperature.

What is an example of pressure?

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7 0
1 year 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.

Now we have to calculate the moles of Na_3AlF_6

From the reaction, we conclude that

As, 1 mole of Al_2O_3 react to give 2 mole of Na_3AlF_6

So, 129.54 moles of Al_2O_3 react to give 129.54\times 2=259.08 moles of Na_3AlF_6

Now we have to calculate the mass of Na_3AlF_6

\text{ Mass of }Na_3AlF_6=\text{ Moles of }Na_3AlF_6\times \text{ Molar mass of }Na_3AlF_6

\text{ Mass of }Na_3AlF_6=(259.08moles)\times (209.9g/mole)=54380.892g=54.38kg

Thus, the mass of cryolite produced will be, 54.38 kg

5 0
3 years ago
2H2O2 →2H2O +O2 ∆H=?<br><br> 2H2 +O2 → 2H2O ∆H=-572 KJ<br><br> 2H2O2 → 2H2 + 2O2 ∆H=376 KJ
Marat540 [252]

Answer:

WOT in the world is this?

Explanation:

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