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>
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Answer:
1) positive
2) carbocation
3) most stable
4) faster
Explanation:
A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.
The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.
Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.
Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.
Because their species is in danger of going extinct<span />
Answer:
transition metals
Explanation:
they're the elements in yellow in the picture
