The volume that will be occupied at 735 torr and 57 c is 23.12 L
<u><em>calculation</em></u>
- <u><em> </em></u> At STP temperature=273 k and pressure=760 torr
- <u><em> </em></u>by use of combined gas formula
that is P1V1/T1= P2V2/T2
where; P1 =760 torr
T1= 273 K
V1= 18.5 L
P2= 735 torr
T2= 57+273= 330 K
V2=?
- by making V2 the formula of subject
V2= T2P1V1/P2T1
V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L
Alkali Metals (Group 1) elements experience an increase in the vigour of their reaction in water as they go down the group (as the atomic number increase). As such the most reactive Alkali Metal would be
FRANCIUM, which is at the base of Group One.
Quite frankly, you do not want Francium to react with water- that's a huge explosion on your hand.
Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y =
= 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>
I would assume carbon and water if not Im sorry.
Nitrogen in the limiting reactant x