Answer:
x = 4.17
y = 1.86
Explanation:
0.62 = log(x)
x = 10^0.62 = 4.17 ( to the nearest hundredth)
0.62 = ln(y)
y = e^0.62 = 1.86 (to the nearest hundredth)
<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 4.53 V.
Answer:
F = 800 N
Explanation:
Given data:
Mass = 80 Kg
Acceleration = 10 m/s²
Force = ?
Solution:
Formula:
<em>F = m × a
</em>
F = force
m = mass
a = acceleration
Now we will put the values in formula:
<em>F = m × a
</em>
F = 80 kg <em>× </em>10 m/s²
F = 800 kg.m/s²
kg.m/s² = N
F = 800 N
Answer:
n = 0.573mol
Explanation:
PV = nRT => n = PV/RT
P = 1.5atm
V = 8.56L
R = 0.08206Latm/molK
T = 0°C = 273K
n = (1.5atm)(8.56L)/(0.08206Latm/molK)(273K) = 0.573mol
<span>There is a direct correlation between the period number and the energy level for valence electrons. For example, the H and He elements, in period 1, have their outer electrons in the energy level "1". This continues down the rows: all the elements in period 2 have their principal energy level as n = 2, period 3 has n = 3, and so on.</span>
