How many elements are in 2CaCO3 and C8H10N402

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

3 years ago

What sequence of enzymes allows for the synthesis of glycogen from glucose-6-phosphate?

Nutka1998 [239]

Answer:

The answer is (e) : phosphoglucomutase, UDP-glucose pyrophosphorylase, glycogen synthase then amylo-(1,4-1,6)-transglycosylase.

Explanation:

Phosphoglucomutase: Convert glucose-6-phosphate to glucose-1-phosphate.

UDP-glucose pyrophosphorylase: Form UDP-glucose from glucose-1-phosphate.

Glycogen synthase: Add the new glucose from UDP-glucose to the growing glycogen chain.

Amylo-(1,4-1,6)-transglycosylase: This is a branching enzyme, it initiates formation of branches evolving from the main chain.

4 0

2 years ago

Someone help please this is due today!!?

maria [59]

Answer:

I think beaker three will take longer to boil since there is more water present,compared to the other beakers

3 0

2 years ago

Please help, I really don’t understand this!!!

kap26 [50]

Analysing the Question:

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: for every 1 mole of Glucose, we need 6 moles of Oxygen

Moles of Glucose used in the reaction:

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

Mass of Oxygen required:

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen = 32 / 30

Mass of Oxygen = 1.06 grams

5 0

3 years ago

He molecular formula mass of this compound is 180 amu . what are the subscripts in the actual molecular formula?

mezya [45]

I can't actually answer this one if the empirical formula is not given. Luckily, I've found a similar problem from another website. The problem is shown in the picture attached. It shows that the empirical formula is CH₂O. Let's calculate the molar mass of the empirical formula.

Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol

Then, let's divide this to the molar mass of the molecular formula.
Molar mass of M.F/Molar mass of E.F = 180/30 = 6

Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
Actual molecular formula = C₆H₁₂O₆

5 0

3 years ago