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Vinil7 [7]
4 years ago
14

You have four bottles labeled 1, 2, 3, 4. Each bottle has a white powder in it. The identities of the substances are held secret

by the teacher. Mr. Sandstone wants you to try and experiment with the substances to figure out what they are. You add vinegar, a weak acid, to each white substance. The results are in the table provided. Which substance is likely to be baking soda (sodium bicarbonate)?
Chemistry
2 answers:
Dmitry_Shevchenko [17]4 years ago
6 0

Answer:

b

Explanation:

KATRIN_1 [288]4 years ago
4 0

If you are using usatestprep its bottle 2

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Mr. Peters’ class conducted a fun experiment during class. For homework, Jason had to analyze the data and make a conclusion. Wh
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B.Several chemical changes occurred. I conclude that a chemical reaction took place

Explanation:

From the observation, Jason can conclude that several chemical changes occurred and a chemical reaction has taken place. In a chemical change, often times, new products are usually formed.

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Which sample of matter is a mixture? Air, water, ammonia
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I think air is a mixture. Water and ammonia are compounds.

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4 years ago
The outer planets are very large making up 99 percent of the mass of the bodies orbiting the Sun.
Advocard [28]
False I believe .......
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3 years ago
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A certain element consists of two stable isotopes. The first has atomic mass of 7.02 amu and a percent natural abundance of 92.6
statuscvo [17]

<u>Answer:</u> The average atomic mass of the element is 6.95 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

We are given:

Mass of isotope 1 = 7.02 amu

Percentage abundance of isotope 1 = 92.6 %

Fractional abundance of isotope 1 = 0.926

Mass of isotope 2 = 6.02 amu

Percentage abundance of isotope 2 = 7.42 %

Fractional abundance of isotope 2 = 0.0742

Putting values in equation 1, we get:

\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]

\text{Average atomic mass of element}=6.95amu

Hence, the average atomic mass of the element is 6.95 amu

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It's a example of a atom c
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