Answer:
Second.
Explanation:
There are four types or kinds of the laws of thermodynamics, but in this question we are actually considering the second law of thermodynamics.
The second law of thermodynamics, in a simple sense means that energies can flow in an equilibrate manner. It can also be stated as the total entropy of the universe. The second law is normal refer to as Law concerning Entropy.
Thus, filling the gap in the Question above. Note that the word written in capital letters is the missing word.
"The SECOND law of thermodynamics states that when two objects of different temperature are in contact, heat energy will flow from the hotter to the cooler object."
Answer: Look it up on the internet
Explanation: hi just wanted to tell you youre amazing
Answer:
<em>Molar mass of the gas is 0.0961 g/mol</em>
<em></em>
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
=
where
= rate of effusion of hydrogen gas
= rate of effusion of unknown gas
= molar mass of H2 gas
= molar mass of unknown gas
substituting values, we have
=
4.587 =
= /4.587
= 0.31
= = <em>0.0961 g/mol</em>
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
Are there any answer choices?