Answer:
see explaination
Explanation:
Molecular equation;
2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)
Total ionic equation; . Includes all ions ;
6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)
Net ionic equation; remove common ions from total ionic;
2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)
G(2)=2
For this, you can plug in 2 everywhere you see an n. So the equation will read:
g(2)=g(2-1)+2 -> g(2)=g(1)+2. Since we are given g(1)=0, we can plug in 0 where we see g(1). The equation is now. g(2)=0+2. So, g(2)=2.
Explanation:
Answer
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The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0
2× ( Oxidation number of Al) +3(−2)=0
2× ( oxidation number of Al) +6
∴ Oxidation number of Al =+3
<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
Solids and liquids have volumes thatdo not change easily. A gas, on the other hand, has a vol- ume thatchanges to match the volume of its container. The molecules in a gas are very far apart compared with the molecules in a solid or a liquid. The amount of space between the molecules in a gas can change easily.
