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Ludmilka [50]
3 years ago
5

Arenediazoniums can undergo electrophilic aromatic substitutions with a wide variety of activated aromatic compounds to yield ne

w azo dyes.
a. True
b. False
Chemistry
1 answer:
mel-nik [20]3 years ago
4 0
B false
I did this last year and it’s false
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How many cm is 0.375 ft.​
ivann1987 [24]

Answer:

11.43 cm

Explanation:

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2 years ago
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A certain reaction is exothermic in the forward direction. The reaction has more moles of gas on the product side. Which of the
alexgriva [62]

Answer:

Decreasing the pressure

6 0
3 years ago
a researcher obtains a sample of 0.070 M nitrate solition. A 20.0 mL aliquote of the nitrate solution is added to 10.0 mL of amm
katovenus [111]

Answer:

Concentration of nitrate in the new solution = 0.007 M

Explanation:

Given:

Concentration nitrate solution = 0.070 m

Volume of aliquote of the nitrate solution is add = 10.0 ml

Total volume = 100 ml

Find:

Concentration of nitrate in the new solution

Computation:

Number of M. mole = 0.070 m x 10.0 ml

Number of M. mole = 0.7 m-moles

Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml

Concentration of nitrate in the new solution = 0.007 M

6 0
2 years ago
Which subatomic particle helps identify an elements identify
Sindrei [870]

Answer:

Protons

Explanation:

Protons are the subatomic particle that determines the identity of an element.

8 0
3 years ago
The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
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