Answer:
V2~0.4839M
Explanation:
We're going to use Boyles law to answer the question.
Boyle's law:
P1V1=P2V2
P1=151mmHg
P2=166mmHg
V1=0.532L
V2=?
V2=(P1 x V1)/P2
V2=(151 x 0.532)/166
V2~0.4839M
Hope it helps:)
The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
Na + NaNO3 = Na2O + N2
4 Na + 2 NaNO3 = 6 Na2O + N2
6 Na on each side
2 N on each side
6 O on each side
The number of years required for 1/4 cobalt-60 to remain after decay is calculated as follows
after one half life 1/2 of the original mass isotope remains
after another half life 1/4 mass of original mass remains
therefore if one half life is 5.3 years then the years required
= 2 x 5.3years = 10.6 years
Answer:
= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl
Explanation:
