Chemistry, 07.03.2020 07:29, cupcake20019peehui

Given the unbalanced equation Ca(s) + Ag+(aq) → Ca2+(aq) + Ag(s), for a reaction, what is being reduced?

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answered: 3ds

Ag^+ is reduced to Ag^0

Explanation:

Based on the equation

Ca+Ag^+=Ca^2+ + Ag

Ca^0 is increased to Ca^2+

And Ag^+ is reduced to Ag^0

answered: BreBreDoeCCx

Ag is reduced in the reaction.

Explanation: From the balanced equation: Ca + Ag+ => Ca2+ + Ag

For an element to be reduced there must be a decrease in oxidation number. For Ca in the reactant it is an independent atom so its oxidation number is 0. The on the product the oxidation number of Ca increases to 2+. So Ca is oxidized.

Ag in the reactant has an oxidation of +1 and decreases on the product side where its oxidation is 0. So Ag is reduced in the reaction since there is a decrease in oxidation number.

answered: Guest

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.

answered: Guest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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Given the unbalanced equation Ca(s) + Ag+(aq) → Ca2+(aq) + Ag(s), for a reaction, what is being redu...