The smallest ionic radius among the following. Li Mg Na+ be+
1 answer:
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Explanation:
The given data is as follows.
Concentration = 0.1
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
=
T = = (30 + 273) K = 303 K
Formula for electric double layer thickness () is as follows.
=
where, = concentration =
Hence, putting the given values into the above equation as follows.
=
=
= m
or, =
= 1 nm (approx)
Also, it is known that =
Hence, we can conclude that addition of 0.1 of KCl in 0.1
of NaBr "
" will decrease but not significantly.
Answer:
A.
B.
Explanation:
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In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:
Thus we proceed as follows:
A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:
Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:
B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:
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