Answer:
Sulphur
Explanation:
as sulphur is also belongs to VI-A group so its donate 2 electron and gained -2 charged
The molarity of the lake water is 0.00001 M and the pH of lake water is 5.
The lake water is acidic.
Explanation:
Data given:
molarity of base solution Mbase = 0.1 M
volume of the base solution Vbase = 0.1 ml or 0.0001 litre
volume of lake water Vlake = 1000ml or 1 litre
molarity of the lake water, Mlake = ?
Using the formula for titration:
Mbase X Vbase = Mlake X
Mlake = ![\frac{Mbase X Vbase }{Vlake}](https://tex.z-dn.net/?f=%5Cfrac%7BMbase%20X%20Vbase%20%7D%7BVlake%7D)
Putting the values in the equation:
Mlake = ![\frac{0.0001 X 0.1}{1}](https://tex.z-dn.net/?f=%5Cfrac%7B0.0001%20X%200.1%7D%7B1%7D)
Mlake = 0.00001 M
The pH of the lake water will be calculated by using the following formula:
pH = -
[
]
pH = -
[ 0.00001]
pH = 5
Answer:
![\boxed{\text{b) reduction potential less than that of oxygen}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7Bb%29%20reduction%20potential%20less%20than%20that%20of%20oxygen%7D%7D)
Explanation:
Electrochemical corrosion of a metal is its spontaneous reaction with oxygen.
Let's compare the standard reduction potentials for a hypothetical metal M with that of oxygen.
<u>E°/V</u>
O₂(g) + 2H₂O(ℓ) + 4e⁻ ⇌ 4OH⁻(aq) 0.4
M²⁺(aq) + 2e⁻ ⇌ M(s) -0.1
To react spontaneously with O₂, the metal must have a ![\boxed{\textbf{reduction potential less than that of oxygen}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctextbf%7Breduction%20potential%20less%20than%20that%20of%20oxygen%7D%7D)
<u>E°/V
</u>
2 × [M(s) ⇌ M²⁺(aq) + 2e⁻ ] 0.1
1 × <u>[O₂(g) +2H₂O(ℓ) + 4e⁻ ⇌ 4OH⁻(aq)] </u> <u>0.4
</u>
2M(s) + O₂(g) + 2H₂O(ℓ) ⇌ 2M²⁺(aq) + 4OH⁻(aq) 0.5
Only then will we get a positive for the overall cell potential and a spontaneous reaction.
<span>Most metals are solid at room temperature with the exception of mercury, which is liquid at room temp. but, most metals turn liquid when the temperature is hot. Like, Gallium is liquid on hot days, but they harden after a while when it's cold.
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