The power used by Alex to drag the log across the yard is determined as 2,656 W.
<h3>Mass of the log</h3>
The mass of the log is calculated as follows;
W = mg
m = W/g
m = (400)/9.8
m = 40.82 kg
<h3>Velocity of the log</h3>
K.E = ¹/₂mv²
v² = 2K.E/m
v² = (2 x 900)/(40.82)
v² = 44.096
v = 6.64 m/s
<h3>Power used by Alex</h3>
P = Fv
P = 400 x 6.64
P = 2,656 W
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The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
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Supposing a temperature of 25 degrees and supposing that all
activity coefficients are 1
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen
ion, would be 10^(-2.50)
pH + pOH = 14
pOH = 14 - pH = 14 - 2.5 = 11.5
MOH- levels would be coordinated with pOH
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12
Therefore, MOH¯ = 3.2 × 10¯12 M
OC. (Third one) is balanced
