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ziro4ka [17]
3 years ago
9

Does the variable increase(1) or decrease(1) under the described conditions?

Chemistry
1 answer:
EastWind [94]3 years ago
5 0

Answer:

Explanation:

Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 5.

Because the volume has decreased, the particles will collide more frequently with the walls of the container. ... When the volume decreases, the pressure increases. This shows that the pressure of a gas is inversely proportional to its volume. This is shown by the following equation - which is often called Boyle's law.

The kinetic energy of the gas molecules increases, so collisions with the walls of the container are now more forceful than they were before. As a result, the pressure of the gas doubles. Decreasing the temperature would have the opposite effect, and the pressure of an enclosed gas would decrease.

For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. That means that, for example, if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.

Temperature, pressure, volume and the amount of a gas influence its pressure.

Gay Lussac's Law - states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure.

i really hope some of this helped i would put more but its a lot too type

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Multiple choice.<br> Please help.
Ludmilka [50]

Answer:

7. B

8. B

9. A

10. A

11. Increases

Again..I hope it helps.✌

7 0
3 years ago
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the
Damm [24]

Answer:

See solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:

\%  M=\frac{m_M}{m_M+2*m_O}*100 \%\\\\59.93\%  =\frac{m_M}{m_M+32.00}*100 \%

Thus, we solve for the molar mass of the metal to obtain:

59.93\% (m_M+32.00) =m_M*100 \%\\\\m_M*59.93\% +1917.76\% =m_M*100 \%\\\\m_M=47.86g/mol

For the subsequent problems, we proceed as follows:

a.

4.00gO_2*\frac{1molO_2}{32.00gO_2}=0.125molO_2

b.

0.400molH_2S*\frac{2molH}{1molH_2S}*\frac{6.022x10^{23}atomsH}{1molH}=4.82x10^{23}atomsH

c.

0.235gNH_3*\frac{1molNH_3}{17.03gNH_3} *\frac{3molH}{1molNH_3}*\frac{6.022x10^{23}atomsH}{1molH}=2.49x10^{22}atomsH

Regards!

7 0
3 years ago
Which is a type of star system?
Pepsi [2]

<h2><em>hope my answer is useful..</em></h2><h2><em>hope my answer is useful.. </em></h2>

7 0
2 years ago
15. The mass of a sample of iron is 114 g, and the density of iron is 7.86 g/cm3. What is the volume of this iron sample?
anyanavicka [17]

Answer:

The answer is

<h2>14.50 cm³</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass of iron = 114 g

density = 7.86 g/cm³

The volume is

volume =  \frac{114}{7.86}  \\  = 14.50381679

We have the final answer as

<h3>14.50 cm³</h3>

Hope this helps you

5 0
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.

6 0
3 years ago
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