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netineya [11]
3 years ago
10

What is one reason that scientists who study weather need extremely advanced technology?

Chemistry
2 answers:
Anton [14]3 years ago
5 0

Answer: C) To monitor weather event restricted several days in advance.

Explanation:

Dovator [93]3 years ago
4 0
3)to monitor weather events restricted to one region
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3 years ago
A copper wire is made of copper (Cu) atoms. Which of the following best describes the copper atoms?
Evgesh-ka [11]

Answer: Option (A) is the correct answer.

Explanation:

An element is defined as a substance that contains atoms of same type.

As copper wire is a substance that is made up of only atoms of copper. And, all these atoms of copper will be the same. Hence, it is an element.

This means that all the atoms of copper will have same number of protons, neutrons and electrons.

Thus, we can conclude that the statement they are all exactly the same best describes the copper atoms.

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3 years ago
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In the formula 5CaSO4 the total number of sulfur atoms present is a. 1. b. 5. c. 9. d. 20?
podryga [215]
The answer is b . (1*5=5)
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3 years ago
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Are isotopes similar to ions, yes or no?
AysviL [449]

Answer:

Yes

Explanation:

Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. Everything else about them is the same.(If you want more explanation tell me).

5 0
1 year 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:

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

5 0
3 years ago
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