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Tatiana [17]
4 years ago
13

Which process is a sudden movement caused primarily by gravity?

Chemistry
1 answer:
REY [17]4 years ago
5 0
B. Definitley Hope this helps
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Which of the following is not an example of a molecule?
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D. F

Molecules are a group of bonded atoms but Fluorine stands on its own
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4 years ago
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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.

6 0
3 years ago
Determine the number of representative particles in each of the following.
klio [65]

Answer: 1. 1.59 x 10^23 Particles

2. 4.79 x 10^21 particles

3. 2.67 x 10^25 particles

4. 2.12 x 10^23 Particles

Explanation: 1mole of any substance contains 6.02x10^23 Particles

1. 0.264 mol of silver will contain = 0.264 x 6.02x10^23 = 1.59 x 10^23 Particles

2. 7.95 x 10^-3 mol sodium chloride will contain = 7.95 x 10^-3 X 6.02x10^23 = 4.79 x 10^21 particles

3. 44.4 mol carbon dioxide will contain = 44.4 x 6.02x10^23 = 2.67 x 10^25 particles

4. 0.352 mol nitrogen will contain = 0.352 x 6.02x10^23 = 2.12 x 10^23 Particles

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3 years ago
Conservation of mass was discussed in the background. Describe how conservation of mass (actual, not theoretical) could be check
Sedbober [7]

Answer:

Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:

1. Weigh all the equipment and materials required in the experiment before the experiment.

2. Avoid spillage and evaporation during the experiment.

3. Weigh all the equipment and materials after the experiment.

If the mass is conserved then weight from step 1 is equal to weight from step

3 0
3 years ago
Need help with balance equation in chemistry
emmasim [6.3K]
The balance is A Go check other answer I posted
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2 years ago
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