Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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Explanation:
2H2+O2------->2H2O
its yr balanced equation.
hope it helps
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The best way to balance an equation is to balance one atom at a time.
You start with two Au atoms on the left, so you know the coefficient of Au on the right has to be 2. So at first we get,
Au2S3 + H2 --> 2Au + H2S
Then, notice you have 3 sulfur atoms on the left, so you need three on the right.
Our equation becomes
Au2S3 + H2 --> 2Au + 3H2S
Lastly, we now have six hydrogen atoms on the right, and only two on the left, so we assign a three to the H2 on the left
Au2S3 + 3H2 --> 2Au + 3H2S Is the balanced final equation.
Ionization energy increases from left to right in the row and from bottom to top in a column. Also as we get closer to the nucleus it would be harder to take electrons out. B (atomic #5) has 2 layers of electron 2 and 3 atom in each layer. P has 15 so it would be 2,8 and 5 respectively. Ca is 20 so 2,8,8,2 and Zn is 30 and it would be 2,18,8,2.
For energy between second and third ionization we are looking at taking out the 3rd electron. B already has 3 electron in the first layer so its easy to take them all. P has 5 in the last layer so again easy. But when we look at Ca and Zn after the 2nd electron (in the last layer) we should change the layer go one layer inside. So this needs more energy. To pick between Zn and Ca (they are in the same row) I mentioned earlier that in one row as we go to the right ionization energy increases so the answer is Zn.
