Lets name one gas sample as A and other gas sample as B.
we can apply ideal gas law equation for both samples
PV = nRT
P - Pressure of A = Pressure of B
V - volume of A = volume of B
n - number of molecules of both A and B being equal is equivalent to number of moles of A = number of moles of B
R - universal gas constant
Tᵃ - temperature of A
Tᵇ - temperature of B
for gas A
PV = nRTᵃ --1)
for gas B
PV = nRTᵇ ---2)
when we divide both equations
1 = Tᵃ / Tᵇ
Tᵃ = Tᵇ
both temperatures are equal
temperature in Celsius + 273 = temperature in Kelvin
therefore 0 °C = 273 K
the correct answer is
A)
<span>The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 0 </span>°<span>C</span>
Answer:
B
Explanation:
heavier particles, which are the ones that move slowly than the lighter particles have a higher temperature
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Answer:
0.19 g
Explanation:
Step 1: Given data
Volume of hydrogen at standard temperature and pressure (STP): 2.1 L
Step 2: Calculate the moles corresponding to 2.1 L of hydrogen at STP
At STP (273.15 K and 1 atm), 1 mole of hydrogen has a volume of 22.4 L if we treat it as an ideal gas.
2.1 L × 1 mol/22.4 L = 0.094 mol
Step 3: Calculate the mass corresponding to 0.094 moles of hydrogen
The molar mass of hydrogen is 2.02 g/mol.
0.094 mol × 2.02 g/mol = 0.19 g
The density of the rock in g/mL : 14 g/ml
Complete question :
A rock with a mass of 70 grams is dropped into a graduated cylinder with 30 mL of water in it. The water level rose to 35 mL calculate the density of the rock in g/mL
<h3>Further explanation</h3>
Density is a quantity derived from the mass and volume
Density is the ratio of mass per unit volume
Density formula:
ρ = density
m = mass
v = volume
mass of rock = 70 g
Volume of the rock : 35 - 30 = 5 ml
So the density :
