Answer:
C) 0.457
Explanation:
The ratio between O2 and H2O is 1:2 according to the balanced equation. You can find how many moles is O2 by : 5.12/22.4 = 0.22857 ( 1 mole = 22.4 litters)
Moles of H2O will be 0.22857 * 2 = 0.457142.
Therefore answer C)
The answer is 3/4.
The coefficient next to each reactant represents the amount of moles. The compound for oxygen is O2 and the compound for aluminum is 4. We can see that the number next to O2 is 3 and the number next to aluminum is 4.
Answer:
The correct option is: Carbonate ion < Carbon dioxide < Carbon monoxide
Explanation:
Bond energy is defined as the average energy needed to break a chemical covalent bond and signifies the strength of chemical covalent bond.
The bond strength of a covalent bond depends upon the <u>bond length and the bond order.</u>
Carbon monoxide molecule (CO) has two covalent bond and one dative bond. Bond order 2.6
Carbon dioxide (CO₂) has two carbon-oxygen (C-O) double bonds of equal length. Bond order 2.0
Carbonate ion (CO₃²⁻) has three C-O partial double bonds. Bond order 1.5
Also, the bond length is <u>inversely proportional to the bond order and bond strength.</u>
Therefore, <u>order of C-O bond length:</u> Carbon monoxide<Carbon dioxide<Carbonate ion
<u>Order of C-O bond order</u>: Carbonate ion<Carbon dioxide<Carbon monoxide
<u>Order of C-O bond strength or energy</u><u>: Carbonate ion<Carbon dioxide<Carbon monoxide</u>
Answer:
It takes 86 days take to cover half of the lake
Explanation:
In the day #1, the amount of the algae is X,
In the day #2 is 2X
In the day #3 is 2*2*X = X*2²
...
In the day #n the amount of the algae is X*2^(n-1)
Assuming X = 1m³. In the day 87, the area infected was:
1m³*2^(87-1)
7.74x10²⁵m³ is the total area of the lake
the half of this amount is 3.87x10²⁵m³
The time transcurred is:
3.87x10²⁵m³ = 1m³*2^(n-1)
Multiplying for 5 in each side:
ln (3.87x10²⁵) = ln (2^(n-1))
58.9175 = n-1 * 0.6931
85 = n-1
86 = n
<h3>It takes 86 days take to cover half of the lake</h3>
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
<span>specific heat solns = specific heat H2O = 4.18 J/K*C </span>
<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol
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