The pressure needed to change the volume to 24 L is 0.796 atm.
Explanation:
It is known by Boyle's law that the pressure experienced by the gas molecules will be inversely proportional to the volume occupied by the molecules.
So as the initial volume is said to be 22.4 L, consider it as V₁ = 22.4 L. Then the initial pressure is said to be 0.853 atm, so P₁ = 0.853 atm. So we have to determine the new pressure P₂ when the volume is changed to V₂ = 24 L. As there is increase in the volume, the pressure should be decreased due to Boyle's law. Thus, as per Boyle's law, the two pressures and their volumes can be related as
Thus, the pressure gets decreased to 0.796 atm on increase in the volume to 24 L.
So the pressure needed to change the volume to 24 L is 0.796 atm.
Answer : The 
must be administered.
Solution :
As we are given that a vial containing radioactive selenium-75 has an activity of 
.
As, 3.0 mCi radioactive selenium-75 present in 1 ml
So, 2.6 mCi radioactive selenium-75 present in 
Conversion :
Therefore, the must be administered.
Answer:
we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
Explanation:
when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate
Answer:
10.2 g
Explanation:
<em>A mixture of ethyl alcohol (molecular mass = 46.1 u) and water (molecular mass = 18.0 u) contains one mole of molecules.</em>
<em />
The mixture contains 20.0 g of ethyl alcohol. The molar mass of ethyl alcohol is 46.1 g/mol. The moles of ethyl alcohol are:
20.0 g × (1 mol / 46.1 g) = 0.434 mol
The sum of the moles of ethyl alcohol and the moles of water is 1 mole.
n(ethyl alcohol) + n(water) = 1 mol
n(water) = 1 mol - n(ethyl alcohol) = 1 mol - 0.434 mol = 0.566 mol
The molar mass of water is 18.0 g/mol. The mass of water is:
0.566 mol × (18.0 g/mol) = 10.2 g
