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3 years ago

Which of the following properly lists the metals in order from decreasing to increasing ease of oxidation?

2 answers:

6 0

I know it is either B or D, since I know that K will oxidize in water, and Ag will have a tougher time oxidizing. I am going to lean towards D for my answer.

lora16 [44]3 years ago

6 0

Answer is: D. Ag < Cu < Cr < K.

Reactivity series is an empirical progression of a series of metals, arranged by their reactivity from highest to lowest (alkaline metals have highest reactivity and Noble metals lowest reactivity):

K > Ba > Sr > Na > Ca > Mg > Be > Al > Mn > Zn > Cr> Fe > Cd > Co > Ni > Sn > Pb > H(in acids) > Cu > Hg > Ag.

Metal higher in the reactivity series will displace another and it will be oxidized.

For example, potassium is more reactive than silver.

Chemical reaction: K + AgCl → KCl + Ag.

Potassium is oxidized (from oxidation number 0 to +1) and silver is reduced (from oxidation number +1 to 0).

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PLEASE HELP WITH CHEMISTRY !! BRAINLIEST AND 10 POINTS !!

NeTakaya

Answer:

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Explanation:

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3 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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3 years ago

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Ymorist [56]

Answer:

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Explanation:

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2 years ago

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