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Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.75 L of a 0.130 M sucro

Advocard [28]

Answer:125.84g

Explanation:Sucrose is dissacharides an organic compound in the class of carbonhydrate with the chemical formula C11H22O11.molar concentration is given by number of moles/Volume,this implies that moles=molar concentration ×Volume=0.130M×2.75L=0.3575moles.

Furthermore,number of moles=Mass of Sucrose/molecular Mass of Sucrose.

From it's formular C11H22O11, molecular Mass is the addition of the mass number which is 12 for C,2 for H and 16 for oxygen,O.so molecular Mass of Sucrose is (12×11)+(2×22)+(16×11)=352.

So mass =moles ×molecular mass=0.3575moles×352g/moles=125.84g

3 0

3 years ago

Nuclear Chemistry Writing Assignment

pogonyaev

Answer:

21 years

% of parent isotope remaining before safe

6.25%

Explanation:

6 0

3 years ago

Read 2 more answers

If 4.50 g of HCl are reacted with 15.00 g of Caco, according to the following balanced chemical equation, calculate the theoreti

Tom [10]

Answer: HCl is the limiting reactant and the theoretical yield is 2.72 g of CO2. If the actual yield was 2.50 g then, the percent yield is 92.0% when rounding off is done only for the final answer.

Further Explanation:

In order to determine the theoretical yield and the percent yield of CO2, the following steps must be done:

Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form.
Determine the theoretical yield for CO2 when the limiting reactant is used.
Get the percent yield by getting the ratio of the actual yield stated in the problem and the calculated theoretical yield multiplied by 100.

Determining the Limiting Reactant

The Limiting Reactant (LR) will produce fewer moles of the products. To check which of the reactants HCl or CaCO3 is the LR, we do dimensional analysis:

For HCl:

$moles\ CO_{2}\ = (4.50\ g\ HCl)\(\frac{1\ mol\ HCl}{36.46094\ g})( \frac{1\ mol\ CO_{2} }{2\ mol\ HCl}) \\moles\ CO_{2}\ =\ 0.0617098$

For CaCO3:

$moles\ of\ CO_{2}\ = (15.00\ g\ CaCO_{3})\ (\frac{1\ mol\ CaCO_{3} }{100.0869\ g\ CaCO_{3} })\ (\frac{1\ mol\ CO_{2} }{1\ mol\ CaCO_{3} })\\moles\ of\ CO_{2}\ = \ 0.1499$

Since HCl produces fewer moles of CO2, then it is the limiting reactant. We will use the given amount to determine the theoretical yield for CO2.

Determining the Theoretical Yield

From Step 1, we know that 0.0617098 moles of CO2 will be produced. We will just convert this to grams.

$grams\ CO_{2}\ =\ (0.0617098\ mol\ CO_{2}) (\frac{44.01\ g\ CO_{2}}{1\ mol\ CO_{2}})\\grams\ CO_{2}\ =\ 2.71585$

Since the answer only requires 3 significant figures, the final answer is 2.72 grams CO2.

Determining the Percent Yield

Dividing the actual yield by the theoretical yield will give us the percent yield, which is an indicator of how efficient the experiment or the method used was.

From the problem, the actual yield was 2.50 g, hence, the percent yield is:

$percent\ yield\ of\ CO_{2}\ = (\frac{2.50\ g}{2.71585\ g}) (100)\\percent\ yield\ of\ CO_{2}\ = 92.05221$

Rounding off to three significant figures, the percent yield is 92.0%. This suggests that the method used is somewhat efficient in producing CO2.

Learn More

Learn More about Limiting Reactant brainly.com/question/7144022
Learn More about Excess Reactant brainly.com/question/6091457
Learn More about Stoichiometry brainly.com/question/9743981

Keywords: stoichiometry, theoretical yield, actual yield

3 0

3 years ago

Homework . Answered

bija089 [108]

We have that the name (not chemical symbol) of the main group element in period 5 and group 3A is

Metalloid boron (B)

From the Question we are told that

The element belongs to

Period 5 and group 3A

Generally

Group 3A of the periodic table includes the metalloid boron (B), aluminum (Al), indium (In), and thallium (Tl) gallium (Ga),

Period 5 is possessed by the metalloid boron (B) of the Group 3A

For more information on this visit

brainly.com/question/13025901?referrer=searchResults

7 0

3 years ago

Lance wants to separate out the salt from a solution of salt water how can he do this

qwelly [4]

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Answer: Boiling the water

I hope this helped!

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4 0

2 years ago