Answer : The correct option is, (2) Cr (Chromium)
Explanation :
The reactivity series of metal are arranged of the reactivity from the highest to the lowest. Reactivity series is used to determine the products of the single displacement reactions. In the single displacement reaction, the most reactive metal displaces the least reactive metal.
From the given reactivity series we conclude that there are two metal (Mg and Cr) are more reactive metal than the Ni and there are two metal (Pb and Cr) are less reactive metal than the Zn. So, the Cr (Chromium) is the metal which is more active than Ni and less active than Zn.
Hence, the correct option is, (2) Cr
Chemical equations must always balance due to the principles outlined in The Law Of Conservation of Matter. This scientific law states that matter cannot be created out of nothing nor can it be destroyed.
Answer:
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Explanation:
Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V
Answer:
The mean free path = 2.16*10^-6 m
Explanation:
<u>Given:</u>
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
<u>To determine:</u>
The mean free path, λ
<u>Calculation:</u>
The mean free path is related to the collision cross section by the following equation:

where n = number density

Substituting for P, k and T in equation (2) gives:

Next, substituting for n and σ in equation (1) gives:
