0.11 moles of the gas are present in the sample of dry gas.
Explanation:
Data given:
mass of the gas = 2.1025 grams
volume of the gas = 2.850 litres
temperature = 22 degrees (273.15+22) = 295.15 K
Pressure = 740 mm Hg or 0.973 atm
moles of the gas =?
R = 0.08206 atmL/Mole K
From the ideal gas law the number of moles can be calculated in the sample of dry gas. Number of moles will be determined by the pressure exerted, volume and temperature of the gas.
The formula:
PV = nRT
n = 
putting the values in the above equation:
n = 
= 0.11 moles
0.11 moles of the dry gas is present in the sample given.
Answer:
547.7 g of C₆H₁₂O₆
Solution:
The balance chemical equation is as follow,
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
According to equation,
6 moles of O₂ burns = 180.56 g of C₆H₁₂O₆
So,
18.2 moles of O₂ will burn = X g of C₆H₁₂O₆
Solving for X,
X = (18.2 mol × 180.56 g) ÷ 6 mol
X = 547.7 g of C₆H₁₂O₆
Answer:
λ = 2.8 m
Explanation:
Given data:
Frequency of radio wave = 106.7 ×10⁶ Hz
Wavelength of radio wave = ?
Solution:
Formula:
Speed of wave = frequency × wavelength
speed of wave = 3×10⁸ m/s
by putting values,
3×10⁸ m/s = 106.7 ×10⁶ Hz × λ
Hz = s⁻¹
λ = 3×10⁸ m/s / 106.7 ×10⁶ Hz
λ = 3×10⁸ m/s / 106.7 ×10⁶ s⁻¹
λ = 0.028×10² m
λ = 2.8 m
