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3 years ago

A reaction that realeases energy in the form of heat or light.

Chemistry

2 answers:

scoray [572]3 years ago

5 0

Hope this helped you :)

Phantasy [73]3 years ago

4 0

Exothermic reaction releases energy in the form of heat or light

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Phase change is based on the_________, which states that phase change occurs with changing average kinetic energy. a. Law of con

AleksAgata [21]

Answer:

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2 years ago

A buffer is formed by mixing 100 ml of 0.1 m nh4cl with 50 ml of 0.3 m nh3. to this solution 13 ml of 0.1 m naoh is added. what

Hoochie [10]

I had the same question last week and my answer was C. Try that.

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3 years ago

A solution is made by mixing 33.0 ml of ethanol, C2H6O and 67.0 ml of water. Assuming ideal behavior, what is the vapor pressure

ivann1987 [24]

Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms

Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles 

Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles 

total moles = .57 + 3.72 = 4.29 moles 

Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13

Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87

Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr 

partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr 

Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr

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3 years ago

What are two possible fates of stars larger than 8 sol

attashe74 [19]

Stars born larger than 8 solar masses usually retain enough mass to undergo core collapse, with the resulting shock wave producing a Type Ib supernova (spectra without Hydrogen or Silicon lines, with Helium lines), a Type Ic supernova (without Hydrogen or Helium or Silicon lines) or a Type II supernova

I hope it’s what you need

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3 years ago

Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.

Assoli18 [71]

Answer:

1.68% is ionized

Explanation:

The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]

Where [] are concentrations in equilibrium

In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:

[C₇H₆O₂] = 0.225-X

[C₇H₅O₂⁻] = X

[H₃O⁺] = X

Replacing:

6.46x10⁻⁵ = [X] [X] / [0.225-X]

1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²

1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0

Solving for X:

X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

That means percent ionization (100 times Amount of benzoic acid ionized over the initial concentration of the acid) is:

0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>

8 0

2 years ago