<span>15.4 milligrams
The ideal gas law is
PV = nRT
where
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = Ideal gas constant (8.3144598 L*kPa/(K*mol) )
T = absolute temperature.
So let's determine how many moles of gas has been collected.
Converting temperature from C to K
273.15 + 25 = 298.15 K
Converting pressure from mmHg to kPa
753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa
Taking idea gas equation and solving for n
PV = nRT
PV/RT = n
n = PV/RT
Substituting known values
n = PV/RT
n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K)
n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) )
n = 0.007897031 mol
So we have a total of 0.007897031 moles of gas particles.
Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So
24/753 = 0.03187251
The portion of hydrogen is 1 minus the portion of water vapor. So
1 - 0.03187251 = 0.96812749
So the number of moles of hydrogen is
0.96812749 * 0.007897031 mol = 0.007645332 mol
Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight.
Atomic weight hydrogen = 1.00794
Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol
Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g
Rounding to 3 significant figures gives 0.0154 g = 15.4 mg</span>
It would go B. A. E. D. C.
Hope I helped!
Hello!
Bases are defined by Arrhenius as substances which release OH⁻ ions when dissolved in water. NO₂⁻ complies with this definition by the chemical reaction that is shown below:
NO₂⁻(aq) + H₂O (l) HNO₂ (aq) + OH⁻(aq)
Have a nice day!
Answer: The new pressure will be 1.1atm
Explanation:
V1 = V
P1 = 3.16atm
V2 = 3V
P2=?
P1V1 =P2V2
3.16 x V = P2 x 3V
P2 = (3.16 x V) /3V
P2 = 1.1atm
Explanation:
The dimensions of a standard backpack is 30cm x 30cm x 40cm
The mass of an average student is 70 kg
We know that, the density of gold is 19.3 g/cm³.
Let m be the mass of the backpack. So,
An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.
