Freidal craft reaction is the attack of a carbon or carbon chain on aromatic ring with the help of anhydrous AlCl3 to produce alkylated benzene ring.
Only ketone not be able to undergo friedal craft reaction as, it is not aromatic compound whereas all the given reactants are aromatic and gives friedal craft reaction.
Answer:
66 grams
Explanation:
For every 1 gram of NH4NO3 equals 2 grams of water
Answer:
Explanation:
Problem 1
<u>1. Data</u>
<u />
a) P₁ = 3.25atm
b) V₁ = 755mL
c) P₂ = ?
d) V₂ = 1325 mL
r) T = 65ºC
<u>2. Formula</u>
Since the temeperature is constant you can use Boyle's law for idial gases:

<u>3. Solution</u>
Solve, substitute and compute:


Problem 2
<u>1. Data</u>
<u />
a) V₁ = 125 mL
b) P₁ = 548mmHg
c) P₁ = 625mmHg
d) V₂ = ?
<u>2. Formula</u>
You assume that the temperature does not change, and then can use Boyl'es law again.

<u>3. Solution</u>
This time, solve for V₂:

Substitute and compute:

You must round to 3 significant figures:

Problem 3
<u>1. Data</u>
<u />
a) V₁ = 285mL
b) T₁ = 25ºC
c) V₂ = ?
d) T₂ = 35ºC
<u>2. Formula</u>
At constant pressure, Charle's law states that volume and temperature are inversely related:

The temperatures must be in absolute scale.
<u />
<u>3. Solution</u>
a) Convert the temperatures to kelvins:
- T₁ = 25 + 273.15K = 298.15K
- T₂ = 35 + 273.15K = 308.15K
b) Substitute in the formula, solve for V₂, and compute:

You must round to two significant figures: 290 ml
Problem 4
<u>1. Data</u>
<u />
a) P = 865mmHg
b) Convert to atm
<u>2. Formula</u>
You must use a conversion factor.
Divide both sides by 760 mmHg

<u />
<u>3. Solution</u>
Multiply 865 mmHg by the conversion factor:

Closed system. If the system is not closed, matter or energy can escape from the system. an example of this is if you react magnesium and hydrochloric acid in a open system. The H₂ gas is going to escape making it look like some of the mass disappeared . in that same reaction some in an open system will also loose heat to the surrounding which will make it look like less heat was produced.
The average atomic mass of the element is the sum of the products of the percentage abundance of isotope and its mass number. Therefore, for atomic mass equal to 58.933, the most abundant isotope is cobalt-59. Thus, the answer is letter C.