<u>0.219 moles </u><u>moles are present in the flask when the </u><u>pressure </u><u>is 1.10 atm and the temperature is 33˚c.</u>
What is ideal gas constant ?
- The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
- The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
- The combined gas law relates pressure, volume, and temperature of a gas.
We simple use this formula-
The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n = of moles (mol) R = the Ideal Gas Law Constant.
68F = 298.15K
V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters
n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)
n = 0.219 moles
Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.
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So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
that 's what I think
Answer:
<h2>HERE'S YOUR ANSWER </h2>
We are given that the ring is made up of 20 carrat gold. 20 carrat gold means 20 parts of gold and 4 parts of other metal (usually copper or silver). Thus, the percentage of gold in 20 carrat gold will be
= (20 / 24) X 100
= 83.333 %
We can now calculate the mass of gold in the ring. We are given that the ring weighs 300 mg. So, the mass of gold in ring will be
= (83.333 / 100) X 300
= 250 mg
Thus, the mass of gold in 300 mg of the ring is 250 mg.
Now the molar mass of gold is 197 g. This means that there are 6.023 X 1023 atoms of gold in 197g of gold. Thus
197g of gold = 6.023 X 1023 atoms of gold
1g of gold = (6.023 X 1023 ) / 197 atoms of gold
250 mg or 0.25g of gold = (6.023 X 1023 X 0.25) / 197 atoms of gold
= 7.643 X 1020 atoms of gold
<h2>HOPE IT HELPS MAN ☺️</h2>
