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borishaifa [10]
3 years ago
14

What is the pH of a 9.50×10−2 M ammonia solution??

Chemistry
1 answer:
Darina [25.2K]3 years ago
4 0
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

c=9,50*10⁻² mol/L

Ammonia solution is a weak electrolyte.
The dissociation constant of ammonium hydroxide is:
K=1.78*10⁻⁵

[OH⁻]=√(Kc)

pH=14-pOH=14+lg[OH]⁻
pH=14+lg(√(Kc))

pH=14+lg(√(1.78*10⁻⁵*9.50*10⁻²))=8.23
pH=8.23
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A 23.7 g piece of iron at 54.9°C is cooled and releases 338 J of heat. The specific heat of iron is 0.450 J/g°C.
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55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?
ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1},

where

  • V_2 is the final volume,
  • V_1 is the initial volume,
  • T_2 is the final temperature in degrees Kelvins.
  • T_1 is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K.

T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K.

Apply Charles's Law to find the new volume of this gas:

\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L.

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