Answer:
Explanation:P = 1 mol x 30.97 g/mol = 30.97 g N = 2 mol x 14.01 g/mol = 28.02 g. O = 4 mol x ... 5.68 g O2 f. Molecules of water produced? 2.46 g H2O ×. 1 mol H2O. 18.02 g H2O. × ... How many grams of sodium sulfate will be produced by the complete ... 2 mol BF3. 3 mol H2. = 1.65 mol BF3 required. Calculate moles H2 required:.
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
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Answer:
Your answer would be a). 2.0 × 10-9
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Work:
In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.
You would use the ph (4.48) to find the ka for "hbro"
[H+]
=
10^-4.48
=
3.31 x 10^-5 M
=
[BrO-]
or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]
Then you would find ka:
(3.31 x 10^-5)^2/0.55 =2 x 10^-9
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<em>-Julie</em>
The number of Protons.
(which is also the ATOMIC NUMBER)
Answer:
2 KNO₃ + 10 K → 5 K₂O + N₂ (option 1)
Explanation:
1. 2 KNO₃ + 10 K → 5 K₂O + N₂
We have in reactant side:
2 K, 2N, 6O and 10 K. In conclussion, 12 K, 2N and 6 O
We have in product side.
10 K, 5O and 2 N
This equation is unbalanced
2. 2 SO₂ + O₂ → 2 SO₃
In reactant side we have 2 S and 6 O
In product side we have 2 S and 6 O
3. SF₄ + 3 H₂O → H₂SO₃ + 4 HF
In reactant side we have 1 S, 4F, 6 H and 3 O
In product side we have 1 S, 4F, 6 H and 3 O