Answer:
a, d and e. are true.
Explanation:
The reaction that occurs is:
Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl
In ideal conditions, the percent yield of the reaction must be 100%. All explanations about why the student could not collect all precipitate are right:
a. The combined reactants were not stirred before filtering the precipitate. Not stirring could not promote all the reaction. <em>TRUE.</em>
b. The student did not completely dry the precipitate before weighing it. If the student don't dry the precipitate, the mass of precipitate must be higher producing a percent yield > 100%. <em>FALSE.</em>
c. The precipitate was not washed prior to drying. Produce more mass. <em>FALSE.</em>
d. A rubber policeman was not used to scrape precipitate from the beaker. If the student doesn't collect all the precipitate the percent yield could be < 100%.. <em>TRUE.</em>
e. The filter paper was not wetted with water prior to filtering the precipitate. <em>TRUE. </em>If you don't wet the filter paper you can lose a part of precipitate from the walls of this one.
Answer:
5.85 moles of carbon dioxide are created.
Explanation:
The balanced reaction is:
CaCO₃ → CaO + CO₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CaCO₃: 1 mole
- CaO: 1 mole
- CO₂: 1 mole
Then you can apply the following rule of three: if by stoichiometry 1 mole of CaCO₃ produces 1 mole of CO₂, 5.85 moles of CaCO₃ will produce how many moles of CO₂?

moles of CO₂= 5.85
<u><em>5.85 moles of carbon dioxide are created.</em></u>
Answer:
answer is strong base hope it helps
The balanced chemical reaction is expressed as:
M + F2 = MF2
To determine the moles of the element fluorine present in the product, we need to determine the moles of the product formed from the reaction and relate this value to the ratio of the elements in MF2. We do as follows:
moles MF2 produced = 0.600 mol M ( 1 mol MF2 / 1 mol M ) = 0.600 mol MF2
molar mass MF2 = 46.8 g MF2 / 0.6 mol MF2 = 78 g/mol
moles MF2 = 46.8 g ( 1 mol / 78 g ) = 0.6 mol
moles F = 0.6 mol MF2 ( 2 mol F / 1 mol MF2 ) = 1.2 moles F