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Morgarella [4.7K]
3 years ago
9

Aluminum and Oxygen gas will react to form Aluminum Oxide (Al2O3). When the equation is balanced, How many moles of Al2O3 are fo

rmed when a mixture of 0.36 moles Al is ignited?
a. 0.12
b. 0.18
c. 0.28
d. 0.46
Chemistry
1 answer:
Olin [163]3 years ago
5 0
The correct answer is B.) 0.18
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How many grams are needed to make 2 L of solution of 4 M LiNO3?
katen-ka-za [31]

Answer:

552 g of LiNO₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 2 L

Molarity of LiNO₃ = 4 M

Mass of LiNO₃ =?

Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:

Volume of solution = 2 L

Molarity of LiNO₃ = 4 M

Mole of LiNO₃ =?

Molarity = mole /Volume

4 = mole of LiNO₃ / 2

Cross multiply

Mole of LiNO₃ = 4 × 2

Mole of LiNO₃ = 8 moles

Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:

Mole of LiNO₃ = 8 moles

Molar mass of LiNO₃ = 7 + 14 + (16×3)

= 7 + 14 + 48

= 69 g/mol

Mass of LiNO₃ =?

Mole = mass /Molar mass

8 = Molar mass of LiNO₃ /69

Cross multiply

Molar mass of LiNO₃ = 8 × 69

Molar mass of LiNO₃ = 552 g

Thus, 552 g of LiNO₃ is needed to prepare the solution.

6 0
3 years ago
Initially, a 55.0 liter compressible container, holding 2.4 moles of a gas, exerts a pressure of 760 millimeters of mercury at a
Bingel [31]
You would have to use the ideal gas law for this:
PV=nRT
Pressure, Volume, n=moles, R gas constant, Temperature in Kelvin
P=nRT/V
(1.8mol)(62.36)(309K)/43.0L = 805mm Hg 
5 0
3 years ago
Read 2 more answers
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
If the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in
anygoal [31]

Answer:

d. supersaturated.

Explanation:

A solution naturally contains a solute and a solvent. The solute is the solid substance that dissolves in the solvent, which is usually a liquid substance. A solution has a maximum amount of solute that can dissolve in its constituent solvent.

However, when the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in the solution at that temperature, the solution is said to be SUPERSATURATED. This means that the solution contains more than the maximum amount of solute.

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