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expeople1 [14]
3 years ago
11

How many moles of CO2 are produced when 5.00 moles of C2H5OH are burned? Please include the steps.

Chemistry
1 answer:
Burka [1]3 years ago
6 0
Just construct the chemical equation.Knowing that a molecule of ethanol has 2 carbon atoms, 2 molecules of carbon dioxide are produced.So the mol ratio of ethanol and CO2 is 1:2 ,so 10 mols of CO2 is produced.Of course this is under the assumption that it is burnt completely.If It is not burnt completely then thats a different story.
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Fuels cell automobiles use__ gas as a fuel
Ray Of Light [21]

Answer:

hydrogen gas

Explanation:

Fuel cell vehicles use hydrogen gas to power an electric motor. Unlike conventional vehicles which run on gasoline or diesel, fuel cell cars and trucks combine hydrogen and oxygen to produce electricity, which runs a motor.

3 0
3 years ago
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Properties of metalloids tend to fall in between the properties of metals and nonmetal.
fredd [130]
IT A A TRUE ANSWER <3
                         
5 0
3 years ago
What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate
denis23 [38]

1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of  Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

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8 0
3 years ago
You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL
pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

3 0
3 years ago
"What volume of" a 0.300 M BaF2 "solution is needed to prepare" 240.0 mL of a 0.100 M F- solution
zhannawk [14.2K]
Use M x V = M' x V'

0.300 x V = 0.100 x 250

V = .......... ml
6 0
3 years ago
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