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3 years ago

What is the Mr of a substance where 1 mole has a mass of 11 g?

Chemistry

1 answer:

iris [78.8K]3 years ago

6 0

Answer:

A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly

12 grams of carbon-12 (i.e., 6.023 X 1023).

Explanation:

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If an object has a Mass of 11g and a volume fo 13mil, what is its density?

Nimfa-mama [501]

Answer:

$\boxed {\boxed {\sf B. \ d \approx 0.85 \ g/mL}}$

Explanation:

Density is found by dividing the mass by the volume.

$d=\frac{m}{v}$

The mass of the object is 11 grams and the volume is 13 milliliters.

$m= 11 \ g \\v= 13 \ mL$

Substitute the values into the formula.

$d= \frac{11 \ g}{13 \ mL}$

Divide.

$d=0.846153846 \ g/mL$

Round to the nearest hundredth. The 6 in the thousandth place tells us to round the 4 to a 5.

$d \approx 0.85 \ g/mL$

The density is about <u>0.85 grams per milliliter.</u>

8 0

3 years ago

Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp

notka56 [123]

Answer:

0.00 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ 30.07 32.00

2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g: 1.50 11.

2. Calculate the moles of each reactant

$\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}$

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

$\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}$

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

$\text{Moles of CO}_{2} = \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

8 0

3 years ago

!!PLEASE ANSWER!! 24 PTS!!

Thepotemich [5.8K]

1.magnesium

2.magnesium

6 0

4 years ago

An example of kinetic energy being converted into heat energy

sammy [17]

Answer:

if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.

Explanation:

As you say, kinetic energy of large objects can be converted into this thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less than.

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3 years ago

How many neutrons does iodine have

SpyIntel [72]

Answer:

It has 53

Explanation:

hope this helps

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3 years ago

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