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Debora [2.8K]
3 years ago
5

Explain how to solve for percent ionization.

Chemistry
1 answer:
swat323 years ago
4 0
Percent ionization can be computed by dividing the concentration of H+ in equilibrium by the original concentration of the solution multiplied by 100 percent. The concentration in equilibrium can be determined through the equilibrium constant inherent to the solution itself and upon its complete dissociation
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One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2. The
sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

7 0
3 years ago
What happened to the amplitude from wave A to wave B?
dangina [55]
The amplitude doubled
4 0
3 years ago
Read 2 more answers
1
zloy xaker [14]

Answer:

1. They lack a nucleus

Following me..

5 0
3 years ago
SHOW YOUR WORK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
WARRIOR [948]

Answer:

1.375%

Explanation:

Percent Error = measured value - accepted value/ accepted value x 100

Measured Value: 15.78

Accepted Value: 16.00

Work:

\frac{15.78-16.00}{16.00} * 100

\frac{-.22}{16} * 100 = -1.375

You cannot have a negative percentage. Therefore the answer is 1.375%

4 0
3 years ago
What are the ion concentrations in a 0.12 M solution of AlCl3? A) 0.12 M Al3+ ions and 0.040 M Cl- ions B) none of the answers C
KIM [24]

Answer:

D

Explanation:

To answer this question, we will need to write the dissociation equation of aluminum trichloride.

AlCl3 ——-> Al3+ + 3Cl-

It can be seen that when aluminum chloride dissociates, it gives one mole of aluminum ion and three moles of the chloride ion.

From here we can see that the concentration of the aluminum chloride equals that of the aluminum ion while that of the chloride ion is thrice that of the aluminum chloride. This means we simply multiply 0.12M by 3 to get the molarity of the chloride ion while that of the aluminum ion remains the same

5 0
3 years ago
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