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3 years ago

How many moles of oxygen will occupy a volume of 25 l at 1.2 atm and 25 c?

Chemistry

2 answers:

MAVERICK [17]3 years ago

3 0

Use the formula PV=nRT
convert T into kelvin by 25+273
put in the values given and calculate n

DENIUS [597]3 years ago

3 0

You have to use the ideal gas law which is PV=nRT
P: pressure of the gas (in this case 1.2 atm)
V: volume of the gas (in this case 25 L)
T: Temperature of the gas (in this case 25 °C which is 298K)
(You need to convert all temperature values to Kelvin when using gas laws)
n: is the number of moles (in this case the unknown)
R: the gas constant which is 0.0821Latm/molK
rewrite the equation so that n=PV/RT than plug in all of the values.
You should get n=1.2262mol

Therefore the answer is 1.23 moles of oxygen.

I hope this helps. Let me know if anything is unclear.

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How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?

lisov135 [29]

Moles of glucose = Molarity x volume solution
= 4.5 x 1.5
= 6.75 moles.

Hope this helps, have a great day ahead!

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3 years ago

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y (1)

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X (2)

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

X = 35,2g

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0

3 years ago

1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a

Archy [21]

Answer:

Explanation:

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃ + Al → 3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

Al : Ag

1 : 3

0.223 : 3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂ → 2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

HgO : Hg

2 : 2

3.12 : 3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂ + 5O₂ → 2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

N₂O₅ : O₂

2 : 5

12.99 : 5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

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Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

C₆H₆ : CO₂

2 : 12

0.103 : 12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

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jeyben [28]

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2 years ago

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fenix001 [56]

Answer:

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