Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M
Answer:
The amount of ammonia needed is 33.3 g
Therefore, we can also say 1.96 moles of NH₃
Explanation:
The reaction is: 3CuO(s) + 2NH₃(g) → 3H₂O(l) + 3Cu(s) + N₂(g)
If we see stoichiometry, 3 moles of water can be produced by 2 moles of NH₃. We propose this rule of three:
3 moles of water can be produced by 2 mole of ammonia
Then, 2.94 moles of water, must be produced by (2.94 . 2) /3 =1.96 moles of NH₃
If we convert the moles to mass. 1.96 mol . 17 g /1mol = 33.3 g
No need 4 calculatinh
the volume needed is 25.o ml coz it requires the same amt of vol to neutralise koh
The balanced reaction would be as follows:
<span>3NaOH + H3PO4 --> Na3PO4 + 3H20
We are given the concentration and the amount of the reactants to be used. We use these values for the calculations as follows:
</span><span>How many mL of .225 M NaOH will react with 4.568g H3PO4?
4.568 g (1 mol / 98 g) ( 3 mol NaOH / 1 mol H3PO4 ) ( 1 L / .225 mol NaOH ) = 0.621 L or 621 mL of .225 NaOH</span>
Answer:
No.
Step-by-step explanation:
When looking at numbers after the decimal point, think of the numbers as real numbers.
In 1.25, think of the .25 as just 25.
Since this is after the decimal point, it is 25 of a 100.
Now, is 25 closer to 4 or not?
Of course, the number has a great distance from 4.
<em>-kiniwih426</em>
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