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pochemuha
3 years ago
7

A given sample of gas has a volume of 5.20 L at 60.°C and 1.00 atm pressure. Calculate its pressure if the volume is changed to

6.00 L and the temperature to 27°C (assume the amount of gas does not change)
Chemistry
1 answer:
BigorU [14]3 years ago
5 0

Answer:

0.78 atm

Explanation:

Applying general gas equation

PV/T= P'V'/T'................ Equation 1

Where P = initial pressure, T = Initial temperature, V = Initial Volume, P' = Final pressure, V' = Final Volume, T' = Final Temperature.

make P' the subject of the equation

P' = PVT'/TV'.............. Equation 2

From the question,

Given: P = 1.00 atm, V = 5.20 L, T = 60°C = (273+60) = 333K, V' = 6.00 L, T' = 27°C = (27+273)K = 300 K

Substitute these values into equation 2

P' = (1×5.2×300)/(333×6)

P' = 1560/1998

P' = 0.78 atm.

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Answer:

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4 0
3 years ago
A hypothetical element consists of four isotopes. Use the information below to calculate the average atomic volume mass of the e
Keith_Richards [23]

Answer:

104.969 amu.

Explanation:

From the question given above, the following data were obtained:

Isotope A:

Mass of A = 107.977 amu

Abundance (A%) = 0.1620%

Isotope B:

Mass of B = 106.976 amu

Abundance (B%) = 1.568%

Isotope C:

Mass of C = 105.974 amu

Abundance (C%) = 47.14%

Isotope D:

Mass of D = 103.973 amu

Abundance (D%) = 51.13%

Average atomic mass =?

The average atomic mass of the element can be obtained as follow:

Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]

Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]

= 0.175 + 1.677 + 49.956 + 53.161

= 104.969 amu

Therefore, the average atomic mass of the element is 104.969 amu.

8 0
3 years ago
How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?
solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

  • mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

  • mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

7 0
3 years ago
I have a chemistry question that I am needing help on please?​
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8 0
3 years ago
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balu736 [363]

Answer:

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Explanation:

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Hope its right !

3 0
3 years ago
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