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2 years ago

An element that has been broken down and released particles and energy is said to be ____.

Chemistry

2 answers:

rjkz [21]2 years ago

8 0

It is said to be radioactive. Hope this was helpful because I'm learning this in school and I think it's right.

KatRina [158]2 years ago

5 0

Radioactive dating, is the answer
hope it helps (:

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What’s the answer to this ( the proton has a ____ charge and it is found in the ____

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ANSWER:

A proton has a "positive charge" and is found in the nucleus.

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2 years ago

Ascorbic acid, or vitamin C (C6H8O6, molar mass = 176 g/mol), is a naturally occurring organic compound with antioxidant propert

Rus_ich [418]

Answer:

The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.

Explanation:

To determine the mass of ascorbic acid knowing the number of moles we use the following formula:

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of ascorbic acid = 2.82 × 10⁻⁴ × 176

mass of ascorbic acid = 496 × 10⁻⁴ g = 0.0496 g = 49.6 mg

daily requirement of ascorbic acid = 70 - 90 mg

The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.

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3 years ago

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A solution that contains a large amount of salt and a small amount of water is said to be a _______ solution. A. diffused B. uns

Pachacha [2.7K]

Answer:

D. Concentrated

Explanation:

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2 years ago

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Why is oxygen written with a subscript 2 in CO2?

Alexeev081 [22]

Answer:

Explanation:

the answer is b

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3 years ago

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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

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2 years ago