Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
Answer:
7 is the mass number of li and 3 is the atomic number
a 3
b 7
c 3
d 4
Answer:
The volume of helium at 25.0 °C is 60.3 cm³.
Explanation:
In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:
K = °C + 273.15
The initial and final temperatures are:
T₁ = 25.0 + 273.15 = 298.2 K
T₂ = -196.0 + 273.15 = 77.2 K
The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.
a) The total pressure of the system is 1.79 atm
b) The mole fraction and partial pressure of hydrogen is 0.89 and 1.59 atm respectively
c) The mole fraction and the partial pressure of argon is 0.11 and 0.19 atm.
<h3>What is the total pressure?</h3>
We know tat we can be able to obtain the total pressure in the system by the use of the ideal gas equation. We would have from the equation;
PV = nRT
P = pressure
V = volume
n = Number of moles
R = gas constant
T = temperature
Number of moles of hydrogen = 14.2 g/2g = 7.1 moles
Number of moles of Argon = 36.7 g/40 g/mol
= 0.92 moles
Total number of moles = 7.1 moles + 0.92 moles = 8.02 moles
Then;
P = nRT/V
P = 8.02 * 0.082 * 273/100
P = 1.79 atm
Mole fraction of hydrogen = 7.1/8.02 = 0.89
Partial pressure of hydrogen = 0.89 * 1.79 atm
= 1.59 atm
Mole fraction of argon = 0.92 / 8.02
= 0.11
Partial pressure of argon = 0.11 * 1.79 atm
= 0.19 atm
Learn more about partial pressure:brainly.com/question/13199169
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Answer:
B.
Explanation:
the SI unit for density is the kilogram per Cubic meter
