CHEMISTRY COLLEGE

Releases sugar (glucose) into the blood stream to power cells A. Brain B. Heart C. Liver D. Stomach and intestines​

Answers

Answer 1
Answer:

Answer:

C Liver

Explanation:

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What reaction conditions most effectively conver a cabocxylic acid to a methly ester?

Answers

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).  

In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.

What is the maximum mass in grams of NH3 that can be produced by the reaction of of 2.5 g N2 with 2.5 g of H2 via the equation below?N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answers

Answer: The mass of NH_3 produced is, 3.03 grams.

Explanation : Given,

Mass of N_2 = 2.5 g

Mass of H_2 = 2.5 g

Molar mass of N_2 = 28 g/mol

Molar mass of H_2 = 2 g/mol

First we have to calculate the moles of N_2 and H_2.

\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=(2.5g)/(28g/mol)=0.089mol

and,

\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=(2.5g)/(2g/mol)=1.25mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 0.089 moles of N_2 react with 0.089* 3=0.267 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 0.089 mole of N_2 react to give 0.089* 2=0.178 mole of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3* \text{ Molar mass of }NH_3

Molar mass of NH_3 = 17 g/mole

\text{ Mass of }NH_3=(0.178moles)* (17g/mole)=3.03g

Therefore, the mass of NH_3 produced is, 3.03 grams.

Assuming that the distances between the two ions are the same in all cases, which of the following ion pairs has the greatest electrostatic potential energy (i.e., largest in magnitude)? ​Please explain your answer.a.) Na+ - Cl- b.) Na+ - O-2. c.) Al+3 - O-2. d.) Mg+2-O-2 e.) Na- -Mg+2

Answers

Answer:

Correct option: C

Explanation:

As given in the question that distance between two ions are same in all cases hence r is same for all.

potential energy:

P.E =(k* q_(1) * q_(2))/(r)

therefore potential energy depend on the two charge muliplication

so higher the charge multiplication higer will be the potential energy.

magnitude of charge multiplication follow as:

a. 1

b. 2

c. 6

d. 4

e. 2

in option C it is higher

so correct option is C

Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 66.93 % yield?

Answers

Answer:

181.39g of AlCl3 is produced

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu

Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:

Molar mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Summary:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Therefore, 54.81g of Al will react to produce = (54.81 x 267)/54 = 271.01g of AlCl3.

Therefore, the theoretical yield of AlCl3 is 271.01g.

Finally, we shall determine the actual yield of AlCl3 produced from the reaction.

This can be obtain as follow:

Percentage yield of AlCl3 = 66.93%

Theoretical yield of AlCl3 = 271.01g

Actual yield of AlCl3 =?

Percentage yield = Actual yield/Theoretical yield x 100

66.93% = Actual yield /271.01g

Actual yield = 66.93% x 271.01

Actual yield = 66.93/100 x 271.01g

Actual yield = 181.39g.

Therefore, 181.39g of AlCl3 is produced from the reaction.

Rhodium has an atomic radius of 0.1345 nm and density of 12.41 gm/cm3 . Determine whether it has an FCC or BCC crystal structure.

Answers

Answer:

FCC.

Explanation:

Hello,

In this case, since the density is defined as:

\rho =(n*M)/(Vc*N_A)

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

Vc_(BCC)=((4r)/(√(3) ) )^3=((4*0.1345x10^(-7)cm)/(√(3) ) )^3=2.997x10^(-23)cm^3\n\nVc_(FCC)=(2√(2)r)^(3) =(2√(2) *0.1345x10^(-7)cm)^3=5.506x10^(-23)cm^3

Hence, we compute the density for each crystal structure:

\rho _(BCC)=(n_(BCC)*M)/(Vc_(BCC)*N_A)=(2*102.9g/mol)/(2.337x10^(-23)cm^3*6.022x10^(23)/mol) =14.62g/cm^3\n\n\rho _(FCC)=(n_(FCC)*M)/(Vc_(FCC)*N_A)=(4*102.9g/mol)/(5.506x10^(-23)cm^3*6.022x10^(23)/mol) =12.41g/cm^3

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.

Which of the following describes the correct order for relative solubility of minerals in sedimentary rocks? a. Evaporate minerals, quartz, and calcite all have the same relative solubility.
b. Evaporate minerals are more soluble than calcite and quartz.
c. Evaporate minerals are more soluble than quartz and less soluble than calcite.
d. Evaporate minerals are less soluble than quartz and calcite.

Answers

Answer:

Evaporate minerals are more soluble than calcite and quartz.

Explanation:

Evaporate minerals are the water soluble minerals which at higher concentration precipitate out and crystallized forming rocks.

example of chemicals present are:

chlorides and sulphates.

Quartz is silica (very less soluble, or insoluble)

Calcite is calcium carbonate, again an insoluble salt.

Thus

Evaporate minerals are more soluble than calcite and quartz.
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