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What is the mass of 9.45 mol of aluminum oxide (Al2O3)?

Wittaler [7]

Answer:

mass = 963.53 grams

Explanation:

mass = moles x molar mass

m = 9.45 x 101.96 (molar mass of Al2O3)

m = 963.53g

3 0

2 years ago

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When 8.0 g H₂ react with 8.0 g O₂ in the reaction 2H₂ + O₂ → 2H₂O, what are the theoretical yield and the limiting reactant?

True [87]

Answer:

Now, we have to determine the limiting reagent.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g of

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.But 29 g is the given amount. Amount of oxygen unreacted = 29 - 24 = 5 g

7 0

3 years ago

An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about

Lisa [10]

Answer:

$C_2H_6$

Explanation:

Hello.

In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:

$m_C=10.2gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1mol C}{1molCO_2}=0.232 molC$

Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:

$m_H=6.3gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH_2}{1molH_2O} =0.7molH$

Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:

$C:\frac{0.232}{0.232}=1\\ \\H:\frac{0.7}{0.232}=3$

Whose molar mass is:

$M_{CH_3}=12+1*3=15g/mol$

Thus, the ratio of the molecular formula to the empirical formula is:

$\frac{30}{15}=2$

Therefore, the molecular formula is twice the empirical formula:

$C_2H_6$

Which is actually ethane.

Regards.

3 0

3 years ago

What formula do you use to calculated density?

Aleks [24]

Answer:

7 AM was the first to be the best Friend and a friend and I have seen the target market share of a million dollars and Diamond.

8 0

3 years ago

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Please help lots of points

vekshin1

Answer:

The equilibrium vapour pressure of a liquid is the pressure exerted by a vapour in a closed vapour and gas system which is in thermal equilibrium.

It is measured using a simple manometer

Explanation:

When liquid and vapour are in a closed container, the molecules of the liquid and vapour at a certain temperature, will attain thermal equilibrium. this means that there is no net transfer of heat between the bodies. The pressure exerted by the vapour on the liquid at this point is called the equilibrium vapour pressure.

To measure the equilibrium vapour pressure, the instrument used is a manometer. To use this, a sample of the liquid is dropped in a sealed flask connected to a manometer, and the increase in the pressure of the system is monitored as the fluid evaporates. This increase in pressure is known as the vapour pressure.

6 0

3 years ago