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Montano1993 [528]
3 years ago
12

Why do scientists use models to study atoms?

Chemistry
1 answer:
umka2103 [35]3 years ago
3 0
Because atoms are the small and can't be seen by your eyes. It is so small that scientists need to use a model. A model help scientists study things. So, scientists needs to study atoms using models.
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What effects would El Niño most likely have on organisms?
Hunter-Best [27]

Answer:

d) el niño would lead to the death of organisms or populations

Explanation:

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3 years ago
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What is the concentration of H+ in a 2.5 M HCl solution?
Pavel [41]

Answer:

The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M

Explanation:

As HCl is a strong acid and hence a strong electrolyte, it will dissociate as

HCl ⟶ H⁺ + Cl⁻

So, The concentration of H⁺ will be 2.5 M (same as HCl)

Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M

<u>-TheUnknownScientist</u><u> 72</u>

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2 years ago
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Which characteristic of a substance is considered a chemical property?
slavikrds [6]
B. It’s reactivity.
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1 year ago
A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam
Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0
3 years ago
A sample of rubidium contains two different isotopes. 72.15% of the sample
Leona [35]

Answer:

Average atomic mass  = 85.557 amu.

Explanation:

Given data:

Percent abundance of Rb-85 = 72.15%

Percent abundance of Rb-87 = 27.85%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

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Average atomic mass =  6132.75 + 2422.95 / 100

Average atomic mass = 8555.7 / 100

Average atomic mass  = 85.557 amu.

4 0
2 years ago
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