Answer:
The essence including its given problem is outlined in the following segment on the context..
Explanation:
The given values are:
Moles of CO₂,
x = 0.01962
Moles of water,
Compound's mass,
= 0.4647 g
Let the compound's formula will be:
Combustion's general equation will be:
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
⇒ 
Now,
x : y : z = 
= 
= 
= 
So that the empirical formula seems to be "C₃H₆O₂".
Answer:
select all of them except they are the biggest
Answer:
P₂ = 13.9 atm (3 sig. figs.)
Explanation:
The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...
P ∝ 1/V => P = k/V => k = P·V
For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂
Given:
P₁ = 1.31 atm.
V₁ = 5.51 L
P₂ = ?
V₂ = 0.520 L
V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)
Answer:
The answer is explained below
Explanation:
If you add dilute HCl (Hydrochloric Acid) to the solution, and you see fizzing, then it is the carbonate. I would recommend doing this under a fume hood, as HCl has a wicked smell, and can make a few people sick to their stomach (however, you probably won't be using 12M concentrated HCl)
