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Flauer [41]
2 years ago
9

How many moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C? Which variables are giv

en?
Chemistry
2 answers:
elixir [45]2 years ago
7 0
PV = nRT
n = PV/RT
n = (2)(3)/(.0821)(273*25)
n = 6/24.466 = 0.245
liubo4ka [24]2 years ago
5 0

Answer: 0.24 moles

Explanation: Using IDEAL GAS LAW

PV=nRT

where,

P = pressure of the gas= 2.0 atm

V = volume of the gas= 3.00L

T =Temperature of gas=25^0C=(25+273)K=298K

n = number of moles of gas=?

R = Gas constant = 0.0821 Latm/moleK

n=\frac {PV}{RT}=\frac {2.0\times 3.0}{0.0821\times 298}=0.24moles


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Which of the following rules should you follow to balance chemical equations?
Firdavs [7]

Answer:

c. add coefficients as needed

Explanation:

A chemical equation is defined as the equation that shows changes in a chemical reaction. A chemical equation consist of reactant and product, reactant is at left side of the arrow and product is at right side of the arrow.

Reactant => Product

While balancing a chemical equation, the basic rule is to balance the coefficient as required. Coefficient represents the number of molecules and is used at front of a chemical symbol. Change in coefficient helps balance the number of atoms or molecules of the substances on both the sides of the arrow.

Subscripts are never allowed to change because it can change the chemical involved in the reaction.

Hence, the correct answer is "c. add coefficients as needed".

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2 years ago
Question 1 of 30
Ksivusya [100]
The answer will be A
3 0
2 years ago
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If 75 grams of oxygen react, how many grams of aluminum are required?
german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

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