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SashulF [63]
2 years ago
7

Explain what happens to an ionic substance when it dissolves in water. Drag the terms on the left to the appropriate blanks on t

he right to complete the sentences. ResetHelp Ionic compounds dissolve in water and dissociate into the Ionic compounds dissolve in water and dissociate into the blank that make up the compound. For example, NaCl in water will form blank ( a q ) (element of the I group) and blank ( a q ) (element of the VII group). that make up the compound. For example, NaCl in water will form Ionic compounds dissolve in water and dissociate into the blank that make up the compound. For example, NaCl in water will form blank ( a q ) (element of the I group) and blank ( a q ) (element of the VII group).(aq)(element of the I group) and Ionic compounds dissolve in water and dissociate into the blank that make up the compound. For example, NaCl in water will form blank ( a q ) (element of the I group) and blank ( a q ) (element of the VII group).(aq)(element of the VII group).
Chemistry
1 answer:
nignag [31]2 years ago
5 0

Answer:

false

Explanation:

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Microwave ovens are able to cook food because they increase the_______ of water molecules in the food.
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potential energy with the heat given to the food

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The density of methanol is 0.7918g/mL. Calculate the mass of 89.9mL of this liquid.
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The mass is 71.2 g.

Mass = 89.9 mL × \frac{0.7918 g}{1 mL} = 71.2 g
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What type of bond joins the carbon atom to each of the hydrogen atoms? A molecule that consists of four hydrogen atoms and one c
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Answer:

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Identify one disadvantage to each of the following models of electron configuration:
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Answer:

The disadvantages of each of the given model of electron configuration have been mentioned below:

1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.

2). Arrow and line diagrams make the counting of electrons and take up too much space.

3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.

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2 years ago
Read 2 more answers
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
2 years ago
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