Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+
From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3
Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
Answer:
It has to have a problem base and a realistic explanation.
Explanation:
It needs to have enough information for you to be able to come up with an answer and realistic explanation.
Hope I helped :)
The answer for this question is D.
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
Answer: The answer is Z only
I’m assuming the answer choices are
1) x only
2) Z only - the answer
3) X and Y
4) X and Z
Explanation:
