Answer:
If the volume is doubled and the number of molecules is doubled, pressure is unchanged
Explanation:
Step 1: Data given
Temperature = constant
Volume will be doubled
Number of molecules will be doubles
Step 2:
p*V = n*R*T
⇒ gas constant and temperature are constant
Initial pressure = n*R*T / V
Initial pressure = 2*R*T/2
Initial pressure = RT
Final pressure = 4*RT / 4
Final pressure = R*T
If the volume is doubled and the number of molecules is doubled, pressure is unchanged
Mass of KCl= 19.57 g
<h3>Further explanation</h3>
Given
12.6 g of Oxygen
Required
mass of KCl
Solution
Reaction
2KClO3 ⇒ 2KCl + 3O2
mol O2 :
= mass : MW
= 12.6 : 32 g/mol
= 0.39375
From the equation, mol KCl :
= 2/3 x mol O2
= 2/3 x 0.39375
=0.2625
Mass KCl :
= mol x MW
= 0.2625 x 74,5513 g/mol
= 19.57 g
Density = 1.01 g/cm^3 or 1.01 kg/dm^3 or 1010 kg/m^3
Density = mass/volume = 1010 g/1000 cm^3 = 1.01 g/cm^3 = 1.01 kg/dm^3
= 1010 kg/m^3
Answer:
249 L
Explanation:
Step 1: Write the balanced equation
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈
The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol
Step 3: Convert "30.0°C" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 30.0°C + 273.15 = 303.2 K
Step 4: Calculate the volume of carbon dioxide
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm
V = 249 L
