Having thin leaves to keep in more water (like like the pine tree :3)
Initial dimension of the pool:
Length (L)= 50.0 m
Width (W)= 25.0 m
Depth (D) = 2.0 m
Initial Volume of the pool (V1) = L*W*D = 50.0 * 25.0 *2.0 = 2500 m3
When the depth is lowered by 3 cm i.e. 0.03 m, the new depth becomes: 2.0 - 0.03 = 1.97 m
The new volume of the pool (V2) = 50.0*25.0*1.97 = 2462.5 m3
Volume of water to be pumped out = V1-V2 = 2500-2462.5 = 35.5 m3
Now, 1 m3 = 1000 L
therefore, 35.5 m3 == 35.5 *10^3 L
It is given that:
3.80 L of water is pumped out in 1 sec
Therefore, 35.5*10^3 L will be pumped out in:
= 1 s * 35.5*10^3 L/3.80 L = 9.34*10^3 sec
Using the atomic weights, K = 39, O=16 so KO2= 39+32=71. So K=39/71= 0.549 of KO2 and O2=32=0.45KO2 and O2=8g and 0.45KO2=8g so KO2=8/0.45=17.77g. So to check, K=0.549 x 17.77g=9.76 and so K+O2= 9.76+8g=17.76.
HCl:
<span>
m=48,2g
M=36,5g/mol
n = m/M = 48,2g / 36,5g/mol = 1,32mol
1mol : 4mol
MnO</span>₂ + 4HCl ⇒ MnCl₂ + Cl₂ + 2H₂O
0,86mol : 1,32mol
limiting reagent
0,33 will react
HCl is limiting reagent.
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