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frutty [35]
3 years ago
12

73.5 grams C, 4.20 grams H, and 72.3 grams Cl. What is the percent composition of each element in this compound?

Chemistry
1 answer:
Umnica [9.8K]3 years ago
6 0
To find percent composition do "parts over the whole" ---> each individual mass over the total mass times 100 (since it is a percentage). 

total mass ("the whole")= 73.5 + 42.0 + 72.3 grams= 150 grams

1. Carbon %= 73.5 grams/ 150 grams x 100= 49.0%

2. Hydrogen %= 4.20 grams/ 150 grams x 100= 2.80%

3. Chlorine %= 72.3 grams/ 150 grams x 100= 48.2%
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Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c
IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
3 years ago
PLEASE HELP!!!!! I need help with this question!
Ivahew [28]

Answer:

Exothermic Reaction

Explanation:

Its a combustion reaction and they are always exothermic in nature.

8 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
A sample of oxygen gas has a volume of 3.24 L at 29°C. What volume will it occupy at 104°C if the pressure and number of mol are
ohaa [14]

<u>Answer:</u> The final volume of the oxygen gas is 4.04 L

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K

Putting values in above equation, we get:

\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L

Hence, the final volume of the oxygen gas is 4.04 L

8 0
4 years ago
Calculate the pH of a 3.58x10^-9 M Nitric acid (a strong acid) solution
lubasha [3.4K]

Answer:

pH= 8.45

Explanation:

when working with strong accids pH = -log(Concentration)

so -log(3.58e-9) = 8.446

7 0
3 years ago
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