First, we need to determine the half reaction of magnesium. It would be expressed as:
Mg2+ + 2e- = Mg
Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:
4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C
We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.
35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
Hm I’m confused, if you mean what are they weighting, it’s they :)
Answer:
Explanation:
the molecular mass of Na2C6H6O7 =236 g\mole it has a sodium that has 23 g/mole so 7.6 g of Na2C6H6O7 has x of sodium mass
236 g/mole ⇒ 23g/mole
<h2> 7.6 g ⇒ ˣ </h2>
7.6 x 23 ÷ 236 = 74.07×10-2 grams of sodium
<h2 />
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
y-y.......................3y...........y
3y = 1.2
y = 0,4M
Na₃PO₄ -----> 3Na(+) + PO₄(3-)
0,4-0,4..............1,2..........0,4
0.........................1,2..........0,4
C = n/V
n = C×V
n = 0,4×0,65L
n = 0,26 mol Na₃PO₄
mNa₃PO₄: (23×3)+31+(16×4) = 164 g/mol
164g ----- 1 mol
Xg -------- 0,26 mol
X = 164×0,26
X = 42,64g Na₂SO₄
