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kkurt [141]
3 years ago
6

Which law states that the volume of a gas is proportional to the miles of the gas when pressure and temperature are kept constan

t ?
Boyle’s law
Dalton’s law
Charles law
Avogadro’s law
Chemistry
1 answer:
IRINA_888 [86]3 years ago
8 0

Answer:

charles law

Explanation:

a law stating that the volume of an ideal gas at constant pressure is directly proportional to the absolute temperature.

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You can boil or evaporate the water and the salt will be left behind as a solid. If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid. So, I would say maybe oil.

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2 years ago
I need help with this please!!
GenaCL600 [577]

Answer:

Option no 3

Explanation:

Metallic elements aren't usually crumbled in normal air pressure and conditions.

6 0
2 years ago
2. Calculate the density of a metal that occupies 17.75 cm and has a mass of 342.93 g. [D = m/V]
satela [25.4K]

Answer:

19.32

Explanation:

the density is given by the mass over volume

the mass In this case is 342.93 and the volume is 17.75

d=342.93g/17.75cm

=19.32g/cm

i hope this helps

3 0
2 years ago
On the periodic table, the average atomic mass for carbon is 12.01 g. Why is it not exactly 12.00?
KiRa [710]
The average the natural abundances of the various isotopes of carbon to arrive at the fractional mass.

5 0
2 years ago
A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe
Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: V_{1} = 571 mL,       T_{1} = 26^{o}C

(a) T_{2} = 5^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL

(b) T_{2} = 95^{o}F

Convert degree Fahrenheit into degree Cesius as follows.

(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL

(c) T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C

The new volume is calculated as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL

8 0
2 years ago
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