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Ugo [173]
3 years ago
12

Carbon monoxide (CO) gas reacts with oxygen (O2) gas to produce carbon dioxide (CO2) gas. If 1.00 L of carbon monoxide reacts wi

th excess oxygen at standard temperature and pressure, what volume of carbon dioxide is produced?
Chemistry
2 answers:
Lunna [17]3 years ago
7 0
1) Chemical equation:

2CO + O2 ---> 2CO2

2) molar ratios 2 moles CO : 1 mol O2 : 2 moles CO2.

3) When temperature and pressure is kept constant, the molar ratios are equal to the volume ratios.

So, at the same temperature and pressure conditions (standard) you ca state

2 L CO : 1 LO2 : 2 L CO2

=> 2L CO : 2 L CO2 => 1L CO : 1 L CO2.

So, 1 liter of CO2 is produced when 1 liter of CO reacts with excess O2.
mojhsa [17]3 years ago
5 0

Answer:

2CO(g)+O2(g)→2CO2(g)

Explanation:

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For maleic acid, hoocch=chcooh, ka1 = 1.42  10–2 and ka2 = 8.57  10–7 . what is the concentration of maleate ion (–oocch=chcoo
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ANSWER

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3 years ago
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Use molecular structure and intermolecular bonding to describe why bromine has a lower boiling point than water
kotegsom [21]

Answer:

Bromine mollecules are held together by van der waals forces while a water molecule constitutes both van der waals forces and hydrogen bomnding

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3 0
2 years ago
what is the pressure, in atmospheres, of 2.97 mol h2 gas if it has a volume of 73 liters when the temperature is 298 k? 0.50 atm
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P x V = n x R x T

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P x 73 = 72.57492

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hope this helps!




6 0
3 years ago
Is it possible to make new water
alexandr1967 [171]

No, Matter cannot be created nor deastroyed.

5 0
2 years ago
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Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
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