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NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3?
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A. It has a very low pH.
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B. It's dissociation is a reversible reaction.
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C. It has a high H+ concentration.
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D. It will release all of its OH- ions.</h2>
Explanation:
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NH3 is a weak alkali that does not dissociate fully into its solution: It's dissociation is a reversible reaction.
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Reactions are also :
Reversible reaction
A reaction in which products can combine back to give reactants under same given condition .
Example : N₂+H₂-------NH₃
Irreversible reaction
A reaction in which the products cant combine back to give reactants under same set of conditions .
Example : Burning of paper
Answer:
11.39
Explanation:
Given that:


Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:

Thus,


Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)


Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)

x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
Answer:
Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Explanation:
1 mole of any compound contains 6.023 × 10²³ molecules.
molecular weight of NaCl is 23 + 35.5 = 58.5 g.
so, 58.5 grams of NaCl makes 1 mole
⇒ 14.5 g of NaCl =
= 0.248 moles.
⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules
= 1.49 × 10²³ molecules.
And 1 molecule contains 1 Na ion and 1 Cl ion.
⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Aluminum has three oxidation states. The most common one is +3. The other two are +1 and +2. One +3 oxidation state for Aluminum can be found in the compound aluminum oxide, Al2O3.