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olga nikolaevna [1]
3 years ago
11

Please help me it has to be right

Chemistry
1 answer:
taurus [48]3 years ago
8 0

Answer:

The negative from object A must transfer to object B

Explanation:

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g A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of
Lemur [1.5K]

Answer:

The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Explanation:

Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).

The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:

W system= -p*∆V

Where:

  • W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
  • p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
  • ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)

In this case:

  • p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
  • ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)

Replacing:

W system= -1.013*10⁶ Pa* (-0.018 m³)

Solving:

W system= 18234 J

<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>

5 0
2 years ago
An element has two naturally-occurring isotopes. The mass numbers of these isotopes are 113 amu and 115 amu, with natural abunda
DENIUS [597]

Answer: Its average atomic mass is 114.9 amu

Explanation:

Mass of isotope 1 = 113 amu

% abundance of isotope 1 = 5% = \frac{5}{100}=0.05

Mass of isotope 2 = 115 amu

% abundance of isotope 2 = 95% = \frac{95}{100}=0.95

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(113\times 0.05)+(115\times 0.95)]

A=114.9amu

Thus its average atomic mass is 114.9 amu

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3 years ago
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The solubility of a compound with a molecular weight of 233.2 is determined to be 27.9 g in 125.0 g of water at 398 K. Express t
sp2606 [1]

Answer:

0.960 m

Explanation:

Given data

  • Mass of the solute: 27.9 g
  • Molar mass of the solute: 233.2 g/mol
  • Mass of the solvent: 125.0 g = 0.1250 kg

First, we will calculate the moles of solute.

27.9 g × (1 mol/233.2 g) = 0.120 mol

The molality of the compound is:

m = moles of solute / kilograms of solvent

m = 0.120 mol / 0.1250 kg

m = 0.960 m

5 0
2 years ago
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