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2 years ago

Calculate the theoretical yield for the bromination of both stilbenes

Chemistry

1 answer:

podryga [215]2 years ago

4 0

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

You might be interested in

what is molarity of a sodium hydroxide solution made by combining 2.0 L of 0.60 M NaOH with 495 mL of 3.0 M NaOH? Assume the vol

hammer [34]

Answer:

Molarity of the sodium hydroxide solution is 1.443 M/L

Explanation:

Given;

0.60 M concentration of NaOH contains 2.0 L

3.0 M concentration of NaOH contains 495 mL

Molarity is given as concentration of the solute per liters of the solvent.

If the volumes of the two solutions are additive, then;

the total volume of NaOH = 2 L + 0.495 L = 2.495 L

the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M

Molarity of NaOH solution = 3.6 / 2.495

Molarity of NaOH solution = 1.443 M/L

Therefore, molarity of the sodium hydroxide solution is 1.443 M/L

8 0

3 years ago

What is the percent of Cr in Cr2O3

IgorLugansk [536]

The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
= $\frac{103.98}{151.98}$ × 100

= 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.

5 0

3 years ago

Read 2 more answers

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago

A 5.0 L flask contains 0.60 g of 02 at a temperature of 22*C, what is the pressure inside the flask?

Kruka [31]

Answer:

The pressure inside the flask is 0,09 atm

Explanation:

We calculate the number of moles of O2, we convert the Temperature into unit KELVIN and use the formula of the ideal gases to calculate the pressure:

Weight 1 mol of 02= 15,999g x2= 32g

32g---1mol 02

0,60g---x=(0,60gx1mol O2)/32g= 0,019mol O2

T(K)=273+22=295K

PV=nRT

Px5.0l= 0,019mol x 0,082 lxatm/Kxmolx295 K

P=(0,019mol x 0,082 lxatm/Kxmolx295 K)/5.0l

P=0,09atm

4 0

3 years ago

. The dissolving process is exothermic when the energy

Serhud [2]

Answer:

Explanation:

The dissolving process depends on the interaction between solute and solvent (solvation) and the breaking up of the intermolecular bond between solutes. The former is exothermic in nature, while the later is endothermic. Energy is released when solute-solvent particles interact. When this energy exceeds the energy required to break intermolecular bonds between the solute particles, dissolution is exothermic.

8 0

3 years ago