1. Cu + 2AgNO3 ==> 2Ag + Cu(NO3)2 ... balanced equation. This is both a single replacement reaction and an oxidation reduction reaction.
Moles of Cu present = 19.0 g Cu x 1 mole Cu/63.55 g = 0.2990 moles Cu
Moles AgNO3 = 125 g AgNO3 x 1 mole AgNO3/169.9 g = 0.7357 moles AgNO3
Which reactant is limiting? It will be Cu because the mole ratio is 2 AgNO3 to 1 Cu and there is more than enough AgNO3. Thus, amount of Ag formed will depend on moles of Cu (0.2990)
Moles of Ag formed = 0.2990 moles Cu x 2 moles Ag/mole Cu = 0.598 moles Ag
Mass (grams) of Ag formed = 0.598 moles Ag x 107.9 g/mole = 64.52 g = 64.5 g of Ag (3 sig. figs.)
Answer: 6.98
The Relative Atomic Mass of Li = (98*7 + 2*6)/98+2
RAM of Li = (686+12)/100
RAM of Li= 6.98
Answer:
4.31 × 10²
Explanation:
Equation of the reaction;
⇌ 
The ICE Table is shown as follows:
⇌ 
Initial 3.10 2.50 0
Change - x -x + 2x
Equilibrium (3.10 - x) 0.0800 2x
From
;
We can see that 2.50 - x = 0.0800
So; we can solve for x;
x = 2.50 - 0.0800
x = 2.42
which = (3.10 -x) will be :
= 3.10 - 2.42
= 0.68
= 2x
= 2 (2.42)
= 4.84
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430.62
≅ 431
= 4.31 × 10²
6.022 × 10^23 × 1.45 × 10^24 = 8.7319 × 10^47