Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Answer:
r = 3.61x
M/s
Explanation:
The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.
r = k.![[S2O2^{-8} ]^{x} x [I^{-} ]^{y}](https://tex.z-dn.net/?f=%5BS2O2%5E%7B-8%7D%20%5D%5E%7Bx%7D%20x%20%5BI%5E%7B-%7D%20%5D%5E%7By%7D)
K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :



x = 1
Now, to find the coefficient y let's do the same for the experiments 1 and 3:




y = 1
Now, we need to calculate the constant k in whatever experiment. Using the first :


k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]
Using the data given,
r = 
r = 3.61x
M/s
Hope this helps
Answer- 1 mole
Answer:
can you provide a picture?