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>
Answer:
0.6743 M
Explanation:
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:
- Molarity = moles / volume
- moles = Molarity * volume
- 0.4293 M * 39.27 mL = 16.86 mmol NaOH
<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.
Finally we <u>calculate the concentration (molarity) of acetic acid</u>:
- 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
Answer:
Carbon dioxide reacts with calcium hydroxide solution to produce a white precipitate of calcium carbonate
Explanation:
. Limewater is a solution of calcium hydroxide. If carbon dioxide is bubbled through limewater, the limewater turns milky or cloudy white
Answer:
6 is the right answer I know cause I like science
Answer:
Se detailed explanation.
Explanation:
Hello,
In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:
- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.
- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.
- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.
- Their electron configurations are:

It means that the both of them are at the
region since it is the last subshell at which their electrons are.
Best regards.