All categories

3 years ago

How many moles of nitrogen are in 3.7 moles of C8H11NO2?

Chemistry

1 answer:

Phoenix [80]3 years ago

4 0

<h3>Answer:</h3>

3.7 Moles of Nitrogen

<h3>Explanation:</h3>

On observing the chemical formula C₈H₁₁NO₂ (might be formula of Dopamine) it is found that one mole of this compound contains;

8 Moles of Carbon

11 Moles of hydrogen

1 Mole of Nitrogen and

2 Moles of Oxygen respectively.

Calculate Number of Moles of Nitrogen:

As,

1 Mole of C₈H₁₁NO₂ contains = 1 Mole of Nitrogen

So,

3.7 Moles of C₈H₁₁NO₂ will contain = X Moles of Nitrogen

Solving for X,

X = (3.7 Moles × 1 Mole) ÷ 1 Mole

X = 3.7 Moles of Nitrogen

You might be interested in

Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t

chubhunter [2.5K]

Answer:

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

Nair = 0.0206

PH20 = 0.19999

Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.

6 0

3 years ago

The ΔHsoln is _____.

nydimaria [60]

Sometimes negative sometimes positive, your answer is B!

4 0

3 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

3 years ago

Calculate the osmotic pressure associated with 50.0 g of an enzyme of molecular weight 98 g/mol dissolved in water to give 2600

andrew-mc [135]

Answer:

π = 4,882 atm

Explanation:

To calculate the osmotic pressure (π), the Van´t Hoff equation must be used, which is:

π x V = n x R x T

Where:

• π: Osmotic pressure, which is the difference between the levels of the solution and the pure solvent through a semipermeable membrane, which allows the passage of the solvent but not the solute

• V: Volume of the solution, in liters unit

• n: Number of moles of solute

• R: Constant of ideal gases, equal to 0.08206 L.atm / mol.K

• T: Absolute temperature, in Kelvin degrees

With the data you provide you can calculate the osmotic pressure by clearing it from the equation, we would be equal to:

π = (n x R x T) / V

However, all data must first be converted to the corresponding units in order to replace the values in the equation.

Solution volume ⇒ go from mL to L: 

1000 mL of solution ____ 1 L

2600 mL of solution _____ X = 2.6 L

Calculation: 2600 mL x 1 L / 1000 mL = 2.6 L

Temperature ⇒ Go from ° C to K 

T (K) = t (° C) + 273.15 = 30.0 ° C + 273.15 = 303.15 K

Number of moles of solute ⇒ It can be calculated since we have the mass of the enzyme and its molecular mass:

98.0 g of enzyme ____ 1 mol

50.0 g of enzyme _____ X = 0.510 moles

Calculation: 50.0 g x 1 mol / 98.0 g = 0.510 moles

Now, you can replace the values in the Van´t Hoff equation and you will get the result:

π = (n x R x T) / V

π = (0.510 mol x 0.08206 L.atm / mol.K x 303.15 K) / 2.6 L = 4.882 atm

Therefore, the osmotic pressure will be 4,882 atm

3 0

3 years ago

Leya [2.2K]

Answer:

V₂ = 530.5 mL

Explanation:

Given data:

Initial temperature = 20.0°C

Final temperature = 40.0 °C

Final volume = 585 mL

Initial volume = ?

Solution:

Initial temperature = 20.0°C (20+273 = 293 K)

Final temperature = 40.0 °C (40+273 = 323 K)

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₁ = V₂T₁ /T₂

V₂ = 585 mL × 293 K / 323 K

V₂ = 171405 mL.K / 323 K

V₂ = 530.5 mL

6 0

3 years ago