Option B is correct
K = Kp /Kr
The given equation indicating, the product containing 6 moles of proton whereas the reactant contains 2 mole of bismuth and 3 mole of hydrogen sulphide.
Hence, in reaction B there are 2 mole of bismuth and 3 mole of hydrogen sulphide reacting to produce 6 moles of proton. whereas the concentration of Bi2S3 is not considered as it is present in solid phase.
<h3>Answer </h3>
After another 5730 years ( three half lives or 17190 years) 17.5 /2 = 8.75mg decays and 8.75g remains left. after three half lives or 17190 years, 8.75 g of C-14 will be
Explanation:
hope this help
Answer: Option (d) is the correct answer.
Explanation:
It is given that molecular formula is 
. Now, we will calculate the degree of unsaturation as follows.
Degree of unsaturation = 
= 
= 9 - 8 + 1
= 2
As the degree of unsaturation comes out to be 2. It means that this compound will contain one ring and one double bond.
Yes, this compound could be an alkyne as for alkyne D.B.E = 2.
But this compound cannot be a cycloalkane because for a cycloalkane D.B.E = 1 which is due to the ring only.
Thus, we can conclude that it is a cycloalkane is not a structural possibility for this hydrocarbon.
Answer:
Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v
Explanation:
(Oxidation) => Mg°(s) => Mg⁺²(aq) + 2e⁻ E°(Mg°/Mg⁺²) = -2.37 v
(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s) E°(Cu⁺²/Cu°) = +0.34 v
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Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)
Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)
= 0.34v + 2.37v = 2.72v
