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DanielleElmas [232]
2 years ago
7

Paintball is a popular recreational activity that uses a metal tank of compressed carbon dioxide or nitrogen to launch small cap

sules of paint. A typical tank has a volume of 508 cubic centimeters. A 340. gram sample of carbon dioxide is added to the tank before it is used for paintball. At 20.oC, this tank contains both CO2(g) and CO2(l). After a paintball game, the tank contains only CO2(g).
Determine the total number of moles of CO2 added to the tank before it is used for paintball.
Chemistry
1 answer:
irinina [24]2 years ago
6 0
We know,
1000 cm³ = 1 L
1 cm³ = 1 / 1000 L
508 cm³ = (1 / 1000) * 508 L = 0.5080 L

So, 508 cm³ = 0.5080 L

Now,
We know,
22.4 L = 1 mole
1 L = 1 / 22.4 mole
0.5080 L = (1 / 22.4) * 0.5080L = 0.0227 moles

So, 0.0277 moles  moles of CO_2 are added to the tank before it is used for paintball.
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Lynna [10]

Answer:

Following are the explanation of the Rube Goldberg device:

Explanation:

According to the Rube Goldberg devices, which conform with "the energy conservation law," choose a chain of events to carry out such a basic task differently, for this unit, a range of instant theatrical resources are converted into possible energy. It is also responsive to an energy conservation law.

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2 years ago
When a solute molecule is solvated, is energy released or absorbed?
Flauer [41]
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2 years ago
How many moles of BCl3 are needed to produce 10.0 g of HCl(aq) in the following reaction? (HCl molar mass is 36.46 g/mol).
Scilla [17]

Answer:

Moles of BCl₃ needed = 0.089 mol

Explanation:

Given data:

Moles of BCl₃ needed = ?

Mass of HCl produced = 10.0 g

Solution:

Chemical equation:

BCl₃ + 3H₂O  →     3HCl + B(OH)₃

Number of moles of HCl:

Number of moles = mass/molar mass

Number of moles = 10.0 g/ 36.46 g/mol

Number of moles = 0.27 mol

Now we will compare the moles of HCl with BCl₃.

               HCl             :           BCl₃

                 3               :             1

             0.27             :            1/3×0.27 = 0.089 mol

6 0
2 years ago
What happened to the volume of gas when the syringe was exposed to various temperature conditions? Using the concepts explored i
rewona [7]

Answer:

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

Explanation:

The expression for Charles's Law is shown below:

\frac {V_1}{T_1}=\frac {V_2}{T_2}

This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.

<u>Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature. </u>

<u>For example , if the temperature of the gas is reduced to half, the volume also reduced to half. </u>

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5 0
2 years ago
If 25.0 g of NH₃ and 45.0g of O₂ react in the following reaction, what is the mass in grams of NO that will be formed? 4 NH₃ (g)
Allisa [31]

Answer:

The correct answer would be : 33.8 g

Explanation:

Molar mass of ammonia,

Molar Mass = 1* Molar Mass(N) + 3* Molar Mass (H)

= 1*14.01 + 3*1.008  = 17.034 g/mol

mass(NH3)= 25.0 g  (given)

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(25.0 g)/(17.034 g/mol)

= 1.468 mol

Now,

Molar mass of O2

= 32 g/mol

mass(O2)= 45.0 g

similar as ammonia

n (O2)=(45.0 g)/(32 g/mol)

= 1.406 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

1.83456 mol of O2 is required  for 1.46765 mol of NH3

by the calculation we have only 1.40625 mol of O2

Thus, the limiting agent will be - O2

now the Molar mass of NO,

= 1*14.01 + 1*16.0

= 30.01 g/mol  (similar formula used for NH3)

Balanced equation :

mol of NO formed = (4/5)* moles of O2

= (4/5)×1.40625  (from above calculation)

= 1.125 mol

mass of NO = number of moles × molar mass

= 1.125*30.01

= 33.8 g

Thus, the correct answer would be : 33.8 g

5 0
2 years ago
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