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3 years ago

Which phenomena naturally warms earth lower atmosphere and surface?

Chemistry

1 answer:

Alenkinab [10]3 years ago

6 0

The phenomenon that naturally warms the earth's lower atmosphere and surface is called the greenhouse effect.

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A device that does work with only one movement is a simple machine <br> true or false

Rainbow [258]

Answer:

true

Explanation:

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2 years ago

0.0200 moles of a compound is found to have a mass of 1.64 g. Find the formula mass of the compound

KatRina [158]

Answer: 82.0 g/mole

Explanation:

Use the units to see that if we divide 1.64 grams by 0.0200 moles, we'll get a number that is grams/mole, the definition of formula mass.

1.64/0.0200 = 82.0 g/mole (3 sig figs)

We can't tell from this alone what the molecular formula might be, but C6H10 (cyclohexene) comes close (82.1 grams/mole).

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2 years ago

Which factor affects a solute's solubility rather than its rate of solution?

Illusion [34]

Answer: A

Explanation:

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2 years ago

Which of the following would have the largest amount of mass?

Elodia [21]

Answer:

1 kg

Explanation:

prefix nano means 10^-9

kilo 10^3

mili 10^-3

centi 10^-2

the largest amount is 1 kg

7 0

2 years ago

Read 2 more answers

If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe

pishuonlain [190]

Answer:

Explanation:

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: 30.01 32.00 46.01

2NO + O₂ ⟶ 2NO₂

Mass/g: 80.00 16.00

2. Calculate the moles of each reactant

$\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}$

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

$\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}$

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

$\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

$\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}$

(b) Mass of NO reacted

$\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}$

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0

2 years ago