Answer:
The answer to your question is 280 g of Mg(NO₃)₂
Explanation:
Data
Efficiency = 30.80 %
Mg(NO₃)₂ = ?
Magnesium = 147.4 g
Copper (II) nitrate = excess
Balanced Reaction
Mg + Cu(NO₃)₂ ⇒ Mg(NO₃)₂ + Cu
Reactants Elements Products
1 Mg 1
1 Cu 1
2 N 2
6 O 6
Process
1.- Calculate the theoretical yield
Molecular weight Mg = 24
Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)
= 24 + 28 + 96
= 148 g
24 g of Mg -------------------- 148 g of Mg(NO₃)₂
147.4 g of Mg ------------------- x
x = (147.4 x 148) / 24
x = 908.96 g of Mg(NO₃)₂
2.- Calculate the Actual yield
yield percent = 
Solve for actual yield
Actual yield = Yield percent x Theoretical yield
Substitution
Actual yield = 
x 908.96
Actual yield = 279.95 ≈ 280g
[OH⁻] = 1.6 × 10⁻⁸ mol / dm³
<h3>Explanation</h3>
By definition, ![[\text{H}^{+}] = 10^{-\text{pH}}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D)
, where ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is the concentration of proton in the solution.
pH = 6.2 for this solution. As a result, 
.
![[\text{H}^{+}] \cdot [\text{OH}^{-}] = \text{K}_w](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Ctext%7BK%7D_w)
, where ![[\text{OH}^{-}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
the concentration of hydroxide ions and 
is the dissociation constant of water.
at 0.10 MPa and 25 °C. As a result, ![[\text{OH}^{-}] = \text{K}_w / [\text{H}^{+}] = 10^{14} / (6.31 \times 10^{-7}) = 1.6 \times 10^{-8} \; \text{mol}\cdot \text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Ctext%7BK%7D_w%20%2F%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B14%7D%20%2F%20%286.31%20%5Ctimes%2010%5E%7B-7%7D%29%20%3D%201.6%20%5Ctimes%2010%5E%7B-8%7D%20%5C%3B%20%5Ctext%7Bmol%7D%5Ccdot%20%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
Answer:
I'm pretty sure it's neutrons the first one
Water , H2O, is a molecule made of oxygen and hydrogen. The bonds that hold water molecules together are due to shared ELECTRONS, and known as covalent bonds
Answer:
3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr
Explanation:
The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.
Here was the process:
CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr
Balances PO4 (2on both sides)
CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr
Balances Lithiums (6 on each side)
3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr
Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)
Hope this helped!
