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tigry1 [53]
3 years ago
7

Identify the solute and the solvent used in this investigation.

Chemistry
1 answer:
Scilla [17]3 years ago
5 0
Solute is the nacl while the solvent is water [H2O]
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What part of a circuit stops the current if it is open
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B is the correct answer

Explanation:

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Balance the reaction. <br> CaO(s) + SO3(g) → CaSO4(s)
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Answer:

CaO + SO 3 → CaSO. 4

This is an acid-base reaction (neutralization): CaO is a base, SO 3 is an acid.

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What is a saturted solution
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Answer:

Please find the definition and further explanation below

Explanation:

Based on the ability for the solvent (liquid substance) to dissolve a solute (solid substance), a solution can either be unsaturated, saturated or supersaturated. A SATURATED SOLUTION is that which contains the maximum amount of solute a solvent can possibly dissolve.

In other words, a saturated solution can no longer dissolve anymore solute, and hence, any further solute added forms crystals or makes the solution supersaturated.

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3 years ago
Which of the following materials is useful for making molds because it has a low melting point?
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3 years ago
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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

4 0
3 years ago
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