To determine the number of moles of air present in the tank, we assume that air is an ideal gas in order for us to use the following equation,
n = PV/RT
Substituting the known values,
n = (195 atm)(350 L) / (0.0821 L.atm/mol.K)(10 + 273.15 K)
n = 2935.91 moles
Thus, the number of moles of air is approximately 2935.91 moles.
So, litmus paper is a qualitative tool for assessing the acidity or basicity of a substance (usually a solution). In general, blue litmus turns red in the presence of an acid, and red litmus turns blue in the presence of a base. They can't really tell you much more than that.
The solutions that are most likely acids are those that turn blue litmus red <em>and </em>do not evoke a color change in red litmus. A solution that turns red litmus blue <em>or </em>does not evoke a color change in blue litmus is likely not an acid. Using these criteria, solutions 4 and 7 are most likely acids since they both turn blue litmus red (and they cause no color change in red litmus).
The correct answer choice would thus be D.
Answer:
The daughter nuclides of these two decay processes are
and
.
Explanation:
The beta emission is represented by:
A = (Z + 1) + (n - 1) = is invariant
n: neutron
p: proton

Hence, the daughter nuclide of the beta emission of Ir-192 is:
Now, electron capture is represented by:
A = (Z - 1) + (n + 1) = is invariant

Then, the daughter nuclide of the electron capture of Ir-192 is:
Therefore, the daughter nuclides of these two decay processes are
and
.
I hope it helps you!