Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
You need to look at the electronegativity and decide wheter the difference of both of the numbers are significant enough to form a polar bond
Answer:
Explanation:
Hello!
In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:
We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:
Therefore, for is to compute the volume of the used base, we proceed as shown below:
And we plug in to obtain:
Best regards!
Answer:
7.3 g (NH₄)₃PO₄
Explanation:
The balanced equation for the reaction is:
H₃PO₄ + 3 NH₃ ----> (NH₄)₃PO₄
To find the mass of ammonium phosphate ((NH₄)₃PO₄) produced, you need to (1) convert grams NH₃ to moles NH₃ (via the molar mass from the periodic table), then (2) convert moles NH₃ to moles (NH₄)₃PO₄ (via mole-to-mole ratio from balanced equation), and then (3) convert moles (NH₄)₃PO₄ to grams (NH₄)₃PO₄ (via molar mass from periodic table). Make sure to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs because the given value (2.5 grams) has 2 sig figs.
Molar Mass (NH₃): 14.01 g/mol + 3(1.008 g/mol)
Molar Mass (NH₃): 17.034 g/mol
Molar Mass ((NH₄)₃PO₄):
3(14.01 g/mol) + 12(1.008 g/mol) + 30.97 g/mol + 4(16.00 g/mol)
Molar Mass ((NH₄)₃PO₄): 149.096 g/mol
2.5 g NH₃ 1 mole NH₃ 1 mole (NH₄)₃PO₄ 149.096 g
--------------- x -------------------- x --------------------------- x --------------------------
17.034 g 3 moles NH₃ 1 mole (NH₄)₃PO₄
= 7.3 g (NH₄)₃PO₄
