Answer:
0.48 V
Explanation:
Usually in the cell notation, the left side shows oxidation. So,
Oxidation half reaction:
Reduction half reaction:
Answer:
The answer is below
Explanation:
Newton's second law of motion states that the force applied to an object is directly proportional to the rate of change of momentum with respect to time, going in the same direction as the force.
Let F = force, m = mass of object, v = velocity of object, mv = momentum.
F = d/dt(mv) = m(dv / dt) = ma; a = acceleration.
Let us assume that the object starts from rest to 5 m/s within 1 seconds, hence:
F = m(dv / dt)
200 N = m[(5 m/s - 0 m/s) / (1 s)]
200 = 5m
m = 40 kg
Answer: 5.66 dm3
Explanation:
Given that:
Volume of neon gas = ?
Temperature T = 35°C
Convert Celsius to Kelvin
(35°C + 273 = 308K)
Pressure P = 0.37 atm
Number of moles N = 0.83 moles
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.37atm x V = 0.83 moles x 0.0082 atm dm3 K-1 mol-1 x 308K
0.37 atm x V = 2.096 atm dm3
V = (2.096 atm dm3 / 0.37atm)
V = 5.66 dm3
Thus, the volume of the neon gas is 5.66 dm3
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
It is approximately 10 ^ -10
