Answer:
Option B is correct = 1,3
Explanation:
Chemical equation:
C₂H₄ + O₂ → CO₂ + H₂O
Balanced chemical equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Step 1:
Left side Right side
C = 2 C = 1
H = 4 O = 3
O = 2 H = 2
Step 2:
C₂H₄ + O₂ → 2CO₂ + H₂O
Left side Right side
C = 2 C = 2
H = 4 O = 5
O = 2 H = 2
Step 3:
C₂H₄ + O₂ → 2CO₂ + 2H₂O
Left side Right side
C = 2 C = 2
H = 4 O = 6
O = 2 H = 4
Step 4:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Left side Right side
C = 2 C = 2
H = 4 O = 6
O = 6 H = 4
Answer: A. Diethyl ether has a very low miscibility in wate
The fact that the diethyl ether is miscible or not in water <u>does not imply a ris</u>k for the person who is working with this reagent in the laboratory.
However, the fact that diethyl ether forms explosive peroxides and that it is highly flammable implies that there is a risk of explosion when exposed to air and sunlight. On the other hand, as option C mentions, if a person inhales a large quantity of this reagent, they may lose consciousness and suffer some injury when fainting, due to the powerful anesthetic effect of this reagent.<u> In conclusion, options B, C and D are statements that imply safety problems associated with the use of diethyl ether in the laboratory, while option A does not imply it.</u>
<u>Answer:</u>
<u>For a:</u> The empirical formula of the compound is 
<u>For b:</u> The empirical formula of the compound is 
<u>Explanation:</u>
We are given:
Percentage of P = 43.6 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.
For Phosphorus = 
For Oxygen = 
Converting the moles in whole number ratio by multiplying it by '2', we get:
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of P : O = 2 : 5
Hence, the empirical formula for the given compound is 
We are given:
Percentage of K = 28.7 %
Percentage of H = 1.5 %
Percentage of P = 22.8 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of K = 28.7 g
Mass of H = 1.5 g
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Potassium =
Moles of Hydrogen =
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.
For Potassium = 
For Hydrogen = 
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of K : H : P : O = 1 : 2 : 1 : 4
Hence, the empirical formula for the given compound is 
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