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Dima020 [189]
3 years ago
5

Which of the following describes a compound?

Chemistry
1 answer:
Gemiola [76]3 years ago
8 0

A compound is two or more chemically combined elements. Fore example NaCl, CO2, H2O

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Chemists recognize many different elements, such as gold, or oxygen, or carbon. Suppose you got some carbon, and started splitti
Otrada [13]

Answer:

An Atom.

Explanation:

An atom is the smallest particle of a chemical element that can exist. It the fundamental piece of matter. Therefore, splitting carbon into smaller pieces , the smallest piece that would still be called "carbon" would be an Atom of carbon. However, atom is also made of three subatomic particles called electrons, protons and neutrons.

6 0
3 years ago
Calculate the rate speed per drop in minute when applying a drip of 1000mg to last 4 2 hours​
Mnenie [13.5K]

Answer:

42 hours is 2520 minutes...

If 1000 mg drops in 2520 minutes,

Then in 1 minute, (1000/2520) drop drops...

Which is approximately 0.40 drops.

4 0
3 years ago
A
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In my opinion the answer is identical
3 0
2 years ago
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A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What i
spayn [35]

Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm

8 0
3 years ago
How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?
PIT_PIT [208]
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
7 0
4 years ago
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