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Evgen [1.6K]
2 years ago
10

What are the condition for a gas like carbon (II) oxide to obey the gas laws?​

Chemistry
2 answers:
qwelly [4]2 years ago
8 0

Answer:

The gas must occupy 0 volume at -273°celcius

Kruka [31]2 years ago
3 0

Answer:

I don't know the answer, but I can give options :)

Explanation:

A gas such as carbon monoxide would be most likely to obey the ideal gas law at:

Option 1) High temperatures and low pressure.

Option 2) Low temperatures and high pressure.

Option 3) High temperatures and high pressure.

Option 4) Low temperatures and low pressure.

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As you move to a higher elevation (e.g. in the mountains), the air pressure drops significantly. In turn,
soldi70 [24.7K]
The answer is C ( The evaporation rate of water increases at all temperatures causing an increase in volume.)                           
3 0
3 years ago
What state of matter is Polaris, the north star?
Anni [7]

The state of matter of Polaris, the north star is gas.

<h3>What are stars made of?</h3>

Stars are made up of a mixture of hot gases.

The mixture consists of helium and hydrogen. Hydrogen burns into helium to give starts a shining appearance when observed from a far distance.

Thus, the state of matter of all stars, including the north star, is gas.

More on stars can be found here: brainly.com/question/21521087

#SPJ1

8 0
2 years ago
A gas that was cooled to 200 Kelvin has a volume of 65.8 L. If its initial volume was 132.4 L, what was its initial temperature?
Mandarinka [93]

Answer:

Initial temperature, T1 = 99.4 Kelvin

Explanation:

<u>Given the following data;</u>

  • Initial volume, V1 = 65.8 Litres
  • Final temperature, T2 = 200 Kelvin
  • Final volume, V2 = 132.4 Litres

To find the initial temperature (T1), we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

\frac {V}{T} = K

\frac {V_{1}}{T_{1}} = \frac {V_{2}}{T_{2}}

Making T1 as the subject formula, we have;

T_{1} = \frac {V_{1}T_{2}}{V_{2}}

Substituting the values into the formula, we have;

T_{1} = \frac {65.8 * 200}{132.4}

T_{1} = \frac {13160}{132.4}

<em>Initial temperature, T1 = 99.4 Kelvin</em>

6 0
3 years ago
By the reaction of carbon &amp; oxygen , a mixture of CO &amp;CO2 is obtained. What is the composition by mass of the mixture ob
Hatshy [7]
M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol

m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol

   2C     +     O₂      →    2CO
1 mol    0.625 mol        1 mol
         0.625-0.5=0.125 mol

      2CO    +         O₂       →        2CO₂
0.250 mol       0.125 mol       0.250 mol

n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g

n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
4 0
3 years ago
Consider the combustion of carbon monoxide (CO) in
Julli [10]

Answer:

3.60 mol CO₂

Explanation:

Balanced chemical reaction:

2CO + O₂ ⇒ 2CO₂

The molar ratio between CO₂ and CO is 1:1

2CO₂/2CO = CO₂/CO

Thus, the moles of CO₂ produced from 3.60 moles of CO is 3.60 moles:

(3.60 mol CO)(CO₂/CO) = 3.60 mol CO₂

6 0
3 years ago
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