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LUCKY_DIMON [66]
3 years ago
13

Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass. determine the empirical formula for a comp

ound that is 36.86% n and 63.14% o by mass. no
Chemistry
1 answer:
nexus9112 [7]3 years ago
5 0
<span>Mass of nitrogen = 14.0067
</span> Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 / 14.0067 = 2.63 
<span>And oxygen = 63.14 / 15.9994 = 3.95 <span>
now we have: N----- 2.63 and O----3.95 
by dividing both with the smallest number we get
</span></span> <span>N-------2.63/2.63 = 1<span>
<span>O-------3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
</span></span></span> N= 1 x 2 = 2 And O = 1.5 x 2= 3
<span>So, the empirical formula is N</span>₂O₃.
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A civil engineer chooses to use wooden beams because they will sag before
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B. choosing a material that will show warning before it fails.

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A state property X has a value 89.6 units. It undergoes the following changes, first increase by
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Answer:

B) -4.1 units

Explanation:

According to this question, a state property X has a value 89.6 units. It undergoes the certain changes as follows:

- first increase by 3.6 units

- then increase by another 18.7 units

- then decrease by 12.2 units

- and finally attains a value of 85.5 units

This can be mathematically represented by 89.6 - {3.6 + 18.7 - 12.2 - x) = 85.5

To get x, we say;

89.6 + 3.6 = 93.2

93.2 + 18.7 = 111.9

111.9 - 12.2 = 99.7

99.7 - 85.5 = 14.2units.

The changes that occured is represented as follows:

= (3.6 + 18.7) - (12.2 + 14.2)

= 22.3 - 26.4

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3 years ago
If a material conducts heat easily, its a good _____. <br> It has to be a 18 letter word
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Answer:

a thermal conductor

Explanation:

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Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
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Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

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3 years ago
I need help identifying the type of reactions. Can anyone help?
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Reaction 8: Combustion reaction.

Explanation:

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<u><em>Reaction 6:</em></u> Zn + 2HCl → H₂ + ZnCl₂.

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  • It is a combustion reaction.
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  • So, it is considered as a combustion reaction.
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