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3 years ago

Which statement is correct about the reaction that occurs between copper and aqueous silver nitrate?

Chemistry

2 answers:

svetlana [45]3 years ago

5 0

Answer:

It is a single replacement reaction because copper replaces silver within a compound as copper is higher than silver in the activity series.

Explanation:

A single replacement reaction occurs between an element and a compound, therefore since copper is an element and sliver nitrate is compound, it is single replacement. This reaction occurs because copper is higher in the activity series then silver.

german3 years ago

3 0

Answer: It is a single replacement reaction because copper replaces silver within a compound as copper is higher than silver in the activity series.

Explanation:

Double displacement reaction is one in which exchange of ions take place.

Example: $2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)$

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

Example: $Cu+2AgNO_3(aq)\rightarrow Cu(NO_3)_2(aq)+2Ag$

Thus copper being more reactive than silver will replace silver from its aqueous solution to give silver and copper nitrate.

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4 years ago

Which aqueous solution will have the highest boiling point temperature? A. 0.100 molal NiBr2(aq) B. 0.250 molal CH3OH(aq) C. 0.1

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Answer: 0.150 m $Na_2SO_4(aq)$ will have highest boiling point.

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

where

$\Delta T_b=T_b-T^o_b$ = elevation in boiling point

[tex[k_b[/tex] = boiling point constant

m = molality

i = Van't Hoff factor

A) 0.100 m $NiBr_2$

i = 3 as $NiBr_2\rightarrrow Ni^{2+}+2Br^-$

concentration will be $3\times 0.100=0.300$

B) 0.250 m $CH_3OH$

i = 1 as $CH_3OH$ is a non electrolyte

concentration will be $1\times 0.250=0.250$

C) 0.100 molal $MgSO_4(aq)$

i = 2 as $MgSO_4\rightarrrow Mg^{2+}+SO_4^{2-}$

concentration will be $2\times 0.100=0.200$

D. 0.150 molal $Na_2SO_4(aq)$

i = 3 as $Na_2SO_4\rightarrrow 2Na^{+}+SO_4^{2-}$

concentration will be $3\times 0.150=0.450$

E. 0.150 molal $NH_4NO_3(aq)$

i = 2 as $NH_4NO_3\rightarrrow NH_4^{+}+NO_3^{-}$

concentration will be $2\times 0.150=0.300$

The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m $Na_2SO_4(aq)$ will have highest boiling point.

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3 years ago