All categories

3 years ago

How many moles of al would be produced from 10 moles of al2o3? (not balanced al2o3 ? al + o2?

Chemistry

1 answer:

Alenkinab [10]3 years ago

3 0

20 moles of Al would be produced from the balanced equation and 10 moles of Al2O3

You might be interested in

How many structural and geometrical isomers are there of chloropropene?

n200080 [17]

I think there might be just 4 or 5 chloropropene.

5 0

3 years ago

How many carbon atoms are in 26-hydrogen alkyne

navik [9.2K]

An alkyne contains four carbon atoms.... so if you do 26 multiplied by 4 it equals 104... I do not know if that’s the answer so I apologize if it’s wrong :,)

5 0

3 years ago

any time an energy conversion takes place, some of the original energy is converted to which one of the following?

Charra [1.4K]

Answer:

Thermal energy.

Explanation:

Anytime energy transfers between 2 places or things some of it is 'wasted' as thermal energy (heat). This could be by friction or a change in temperature from a reaction. We say it is wasted because this heat energy serves no purpose.

3 0

2 years ago

I need to know number 2 3 4 and 5

dexar [7]

Answer 2: 57.21235000000001

Answer 3: 44.01g

Answer 4: 24.645320000000005

Answer 5: 107.04578240000001

Good luck

7 0

2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

3 years ago