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2 years ago

What does it mean for a reaction to release energy?.

2 answers:

7 0

Chemical reactions that release energy are called exothermic. In exothermic reactions, more energy is released when the bonds are formed in the products than is used to break the bonds in the reactants. Exothermic reactions are accompanied by an increase in temperature of the reaction mixture.

8090 [49]2 years ago

4 0

Answer:

exothermic

Explanation:

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Heat Transfer Lab

scoray [572]

Heat Transfer Lab

The following represents a lab set up for heat transfer. The cup on the left started with boiling water at 100 degrees C and the cup on the right has water at 20 degrees C. There is an aluminum bar between the two cups allowing heat to transfer from one cup into the other. The set up will be left alone for 20 minutes and temperatures of each cup of water will be recorded every minute for 20 minutes.

mag-aral ka

5 0

2 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

A student was assigned the task of determining the identity of an unknown liquid. The student weighed a clean, dry 250-mL Erlenm

LiRa [457]

Answer:

248.4 mL

Explanation:

Erlenmeyer = 78.649 g

Erlenmeyer + Water = 327.039 g

Water = (Erlenmeyer + Water) - Erlenmeyer

Water = 327.039 - 78.649

Water = 248.4 g

if the density of water is 1 g/mL, we can say that each mL of water weigh 1 g, so we have 248.4 mL of water in the Erlenmeyer Flask.

4 0

3 years ago

What is 4.07 cm x 2.3cm, rounded to the correct number of significant figures

Tom [10]

Answer: 9.4cm^2

Explanation:

8 0

3 years ago

Equal amounts of N2 and O2 are added, under certain conditions, to a closed container. Which changes occur in the reverse reacti

geniusboy [140]

1) The forward reaction is N2 (g) + O2 (g) → 2NO

(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.

2) The equiblibrium equation is

N2 (g) + O2 (g) ⇄ 2NO

3) Then, the reverse reaction is

2NO → N2(g) + O2(g)

Answer: 2NO → N2(g) + O2(g)

4 0

3 years ago