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NeX [460]
3 years ago
7

What is the molar mass of Al(NO3)3 · 3H2O?

Chemistry
1 answer:
Nana76 [90]3 years ago
3 0

Answer:

Go to your periodic table and look for each element. Find the mass for Aluminum, Nitrogen, Oxygen, and Hydrogen.

After you find those, for each compound you must add them together and find the least amount of sig figs after the decimal point.

For the first compound, Al(NO3)3, you will have a total of

1 Aluminum atom

3 Nitrogen atoms

9 Oxygen atoms

Aluminum has 26.982 g/mol

Nitrogen has 14.007 g/mol

Oxygen has 15.999 g/mol

Now multiply those numbers by the amount of atoms of each element.

(1)26.982 g/mol = 26.982 g/mol

(3)14.007 g/mol = 42.021 g/mol

(9)15.999 g/mol = 143.991 g/mol

now add all of those numbers together, and you see that their least significant figure after the decimal is 3, so round to 3 digits after the decimal point.

26.983 + 42.021 + 143.991 = 212.994

now do the same for the other compound

to start you off.. you have

6 Hydrogen atoms

3 Oxygen atoms

Hydrogen has 1.009 g/mol

Oxygen has 15.999 g/mol

the least significant figure after the decimal point is 3, so round you 3 digits after the decimal point.

after you finish getting your totals, you. until them and find the greatest sig fig over all. comment on this if you need further instruction :)

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<h3><u>Answer;</u></h3>

It makes the reaction harder to start

<h3><u>Explanation</u>;</h3>
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  • The source of activation energy needed to push chemical reactions forward is obtained from the surroundings. Catalyst speed up chemical reaction by lowering the activation energy. Therefore, catalysis is the increase in the rate of a chemical reaction by lowering its activation energy.
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D. Photosynthesis uses carbon dioxide while cellular produces carbon dioxide

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If a gas is moved from a large container to a small container but its temperature and number of moles remain the same, what woul
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To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

 

P1V1 =P2V2

<span>P2 = P1V1/V2</span>

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