Yes it can. i hope i helped
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:
The answer to the question is
The rate constant for the reaction is 1.056×10⁻³ M/s
Explanation:
To solve the question, e note that
For a zero order reaction, the rate law is given by
[A] = -k×t + [A]₀
This can be represented by the linear equation y = mx + c
Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀
Therefore the rate constant k which is the gradient is given by
Gradient =
where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M
=
= -0.001056 M/s = -1.056×10⁻³ M/s
Threfore k = 1.056×10⁻³ M/s
Answer:
0.159 M
Explanation:
convert from mL to L then use the equation:
M1V1 = M2V2
rearrange to find M2


0.048 percent is the answer