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in periods the atomic radius decreases
Answer:
1. Zn is oxidize.
Cu is reduced.
2. Fe is oxidize.
H2 is reduced.
Explanation:
1. Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
From the equation,
Zn changes oxidation number from 0 to +2. Therefore Zn is oxidized.
Cu2+ changes oxidation number from +2 to 0. Therefore, Cu2+ is reduced
2. 3Fe(s)+4H2O(g)→Fe2O3(s)+4H2(g)
From the equation,
Fe changes oxidation number from 0 to +3. Therefore, Fe is oxidized.
H changes oxidation number from +1 to 0. Therefore, H is reduced
Answer:
is the formula for the limiting reagent.
Mass of silver chloride produced is 71.8 g.
Explanation:
Moles of silver nitrate = 0.500 mol
Moles of copper(II) chloride = 0.285 mol
According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :
of copper(II) chloride
As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.
And moles of silver chloride to be formed will depend upon silver nitrate.
According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give :
of silver chloride
Mass of silver chloride produced:
0.500 mol × 143.5 g/mol = 71.8 g
Answer:
The 35-watt light bulb does a work of 46200 joules in 1320 seconds.
Explanation:
From Physics we remember the definition of work (), measured in joules, by the following integral equation:
(1)
Where is the power, measured in watts.
If the power is stable in time, then the work can be calculated by the following algebraic equation:
(2)
If we know that and , then the work done by the 35-light bulb is:
The 35-watt light bulb does a work of 46200 joules in 1320 seconds.