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mars1129 [50]
3 years ago
12

Based on the molecular structures of chloromethane and methane shown, a student makes the claim that a pure sample of chlorometh

ane has stronger intermolecular forces than a pure sample of methane has.
Chemistry
1 answer:
goblinko [34]3 years ago
7 0

Answer:

Yes, chloromethane has stronger intermolecular forces than a pure sample of methane has.

Explanation:

In both methane and chloromethane, there are weak dispersion forces. However, in methane, the dispersion forces are the only intermolecular forces present. Also, the lower molar mass of methane means that it has a lower degree of dispersion forces.

For chloromethane, there is in addition to dispersion forces, dipole-dipole interaction arising from the polar C-Cl bond in the molecule. Also the molar mass of chloromethane  is greater than that of methane implying a greater magnitude of dispersion forces in operation.

Therefore, chloromethane has stronger intermolecular forces than a pure sample of methane has.

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Answer:

high B or a low A:) Good Luck

Explanation:

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3 years ago
Consider the three equations below.
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Answer:

Nuclear fusion plays an important role in making elements that are heavier than helium.

Explanation:

Nucleosynthesis is the process by which new atomic nuclei are created from pre-existing nucleons (protons and neutrons) and nuclei. According to current theories, the first nuclei were formed a few minutes after the Big Bang, through nuclear reactions in a process called Big Bang nucleosynthesis.

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Elements heavier than helium are formed by nuclear nucleosynthesis in which nuclear fusion plays a very crucial role as typified by the equations shown in the question.

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3 years ago
Determine the location of the last significant place value by placing a bar over the digit.
Vlad1618 [11]

Answer:

8,040  

0.0300  

699.5  

2.000 x 102

0.90100  

90, 100  

4.7 x 10-8  

10,800,000.0  

3.01 x 1021

0.000410

Explanation:

First remember the following rules of determining the last significant place value :

1. The digits from 1-9 are all significant and zeros between significant digits are also significant.

2.  The trailing or ending zeroes are significant only in case of a decimal number otherwise they are ignored. However starting zeroes of such a number are not significant.

Now observing above rules, lets determine the location of the last significant place value of each given example. I am determining the location by turning the last significant place to bold.

1) 8,040

8,040

Location of the last significant place value is 3 and bar is over last significant digit that is 4. Number is not decimal so ending zero is ignored. Every non zero digit is a significant.

2)  0.0300

0.0300

Location is 3 and bar is over 0. Number has a decimal point so ending zero is not ignored but starting zeroes are ignored.

3) 699.5

699.5

Location is 4 and bar is over 5.

4) 2.000 x 10²

2.000 x 10²

Location is 4 and bar is over 0. This is because the number is decimal so trailing zeroes cannot be ignored. Also if we convert this number it becomes:

200.0 so last significant digit is 0 and location of last significant digit is 4.

5) 0.90100

0.90100

Location is 5 and bar is over 0. This is because in a number with decimal point starting zeroes are ignored but trailing zeroes after decimal point are not ignored. So we count from 9 and last significant digit is 0.

6) 90, 100

90, 100

Location is 3 and bar is over 1. This is because it is not a number with decimal point. So the trailing zeroes are ignored. The count starts from 9 and last significant is 1.

7) 4.7 x 10⁻⁸

4.7 x 10⁻⁸

Location is 2 and bar is over 7. This is because the starting zeroes in a number with a decimal point are ignored. So the first digit considered is 4 and last significant digit is 7. If we expand this number:

4.7 x 10⁻⁸ =    0.000000047 = 0.000000047

Here the starting zeroes are ignored because there is a decimal point in the number.

8) 10,800,000.0

10,800,000.0

Location is 9 and bar is over 0.  Number has a decimal point so ending zero is not ignored and last significant figure is 0.

However if the number is like:

10,800,000. Then location would be 8 and bar is over 0.

9) 3.01 x 10²¹

3.01 x 10²¹

Location is 3 and bar is over 1. Lets expand this number first

3.01 x 10²¹ = 3.01 x 1000000000000000000000

                  = 3010000000000000000000

So this is the number:

3010000000000000000000

Since this is number does not have a decimal point so the trailing zeroes are ignored. Hence the count starts from 3 and the last significant figure is 1

10) 0.000410

0.000410

Location is 3 and bar is over 0. This is because the number has a decimal point so the ending zero is not ignored but the starting zeroes are ignored according to the rules given above. Hence the first significant figure is 4 and last significant figure is 0.

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Answer:

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Explanation:

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Describe the properties of alkaline earth metals. Based on their electronic arrangement, explain whether they can exist alone in
zheka24 [161]

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All the alkaline earth metals readily lose their two outermost electrons to form cations with a 2+ charge.

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Magnesium and calcium are ubiquitous and essential to all known living organisms.

Explanation:

7 0
2 years ago
Read 2 more answers
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