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IgorLugansk [536]

3 years ago

10

Suppose you place an ice cube into a cup of hot tea? What will happen to the temperature of the ice cube and the tea? A) The tem

perature of the ice will increase quickly, while the temperature of the tea will drop slowly. The liquids may or may not reach equilibrium. B) The temperature of the tea will drop quickly, while the temperature of the ice will increase slowly. The liquids may or may not reach equilibrium. C) The ice will melt and the temperature of the tea will drop quickly; then the change will slow until equilibrium is reached. D) The temperature of the ice will rise slowly at first and the temperature of the tea will drop slowly; then the temperature change quicken until equilibrium is reached.

1 answer:

Ksivusya [100]3 years ago

8 0

Answer:

The answer is D

Explanation:

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Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

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

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

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:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

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2 years ago

Why is silicon dioxide a suitable material for lining furnaces

joja [24]

It has a very high melting point

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2 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

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The acid from burning coal is rained back down onto the planet surface and reacts with

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What? Why does it have a question mark it makes no sense because there isn’t any options and you didn’t explain what the question is about.

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Hunter’s velocity was 4.5 m/s. At the end of his race, his velocity was the same. Which best describes his movement?

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If an object is not changing its velocity, then the object is not accelerating. (There’s no acceleration)

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