Answer:
Hydrogen, H_2
Explanation:
mass of each gas is 10.0 g
number of mole = mass/ molar mass
number of moles is directly proportional to volume at constant temp and pressure
this implies that the volume is inversely proportional to molar mass. And Among all the gases in periodic table the molar mass of Hydrogen is the least.
molar mass of H2=2 g/mol
Since, H2 has minimum molar mass then for the same mass of the gases Hydrogen will have maximum volume.
D
Avogadro's number allows us to measure the amount of atoms or molecules in one mole of a substance.
Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.
Answer:
P = 20.1697 atm
Explanation:
In this case we need to use the ideal gas equation which is:
PV = nRT (1)
Where:
P: Pressure (atm)
V: Volume (L)
n: moles
R: universal gas constant (=0.082 L atm / K mol)
T: Temperature
From here, we can solve for pressure:
P = nRT/V (2)
According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.
The volume for any cylinder would be:
V = πr²h (3)
Replacing the data here, we can solve for the volume:
V = π * (17)² * 164
V = 148,898.93 cm³
This volume converted in Liters would be:
V = 148,898.93 mL * 1 L / 1000 mL
V = 148.899 L
Now we can solve for pressure:
P = 125 * 0.082 * 293 / 148.899
<h2>
P = 20.1697 atm</h2>
Answer:
Explanation:
Given:
For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.
To find: fraction of the whole athletic field reserved for each fifth class
Solution:
Fraction of the whole athletic field reserved for four fifth classes = 
So, fraction of the whole athletic field reserved for each fifth class = 
