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2 years ago

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Question 5 What law states that energy can neither be created nor destroyed, but it is transformed? Question 5 options: Law of T

hermonuclear Energy Law of Kinetic Energy Law of Conservation of Energy Law of Thermal Energy

2 answers:

masya89 [10]2 years ago

3 0

It is the law of conservation of energy

zloy xaker [14]2 years ago

3 0

The answer is: LAW OF CONSERVATION OF ENERGY

Hope this helps!! :)

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D = M / V.

Another way to remember the formula for Density is to remember "Mass per unit of volume".

I hope this helps!

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3<br>Nitrogen obtained in the laboratory is collected over water.why?

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NaOH(aq) + MgSO4(aq) →

Cerrena [4.2K]

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2 years ago

You have a 250. -ml sample of 1. 28 m acetic acid (ka = 1. 8 x´ 10–5). calculate the ph of the best buffer.

Nadya [2.5K]

The ph of the best buffer is 4.74

The given acetic acid is a weak acid

The equation of the pH of the buffer

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1 year 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:

50

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

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2 years ago