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2 years ago

The weak acid HZ has a Ka of 2.55x10⁻⁴.(a) Calculate the pH of 0.075 M HZ.

Chemistry

1 answer:

Dima020 [189]2 years ago

4 0

2.37 is the pH of 0.075 M HZ.

The term pH, which originally stood for "potential of hydrogen" (or "power of hydrogen"), is used in chemistry to describe how acidic or basic an aqueous solution is. Lower pH values are summarized for acidic solutions (solutions with higher H+ ion concentrations) than for basic or alkaline solutions.

The pH scale is inversely indicates to the concentration of hydrogen ions in the solution and is logarithmic.

⇒pH = -log( $a_{H+}$ )

Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C (77 °F). At this temperature, solutions with a pH of 7 are neutral (e.g. pure water). The pH neutrality relies on temperature, falling below 7 if the temperature rises above 25 °C.

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3 years ago

A piece of potassium metal is added to water in a beaker. The reaction that takes place is 2K(s) 2H20(/) ..... 2KOH(aq) H2(g) Pr

kakasveta [241]

Answer:

Explanation:

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Gas is evolving so work done by gas is positive or w is positive.

Change in internal energy that is ΔU is negative.

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3 years ago

I need help with this, please :00000

mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if 1 mole of N₂ produce 2 moles of NH₃

then 3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

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