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kotykmax [81]
3 years ago
12

What type of bond would occur between carbon (C) and nitrogen (N)

Chemistry
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

covalent

Explanation:

The carbon and the nitrogen very often form bonds in nature, carbon-nitrogen bonds, which are covalent types of bonds. In fact, the bonds between the carbon and nitrogen are one of the most abundant in the biochemistry and the organic chemistry. The bonds between these two can be double bonds, as well as triple bonds. The carbon-nitrogen bonds have the tendency to be strongly polarized toward the nitrogen.

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Round to 4 significant figures.<br> 0.007062
lapo4ka [179]

Answer:

Hey there!

This is already rounded to four significant figures!

Zeroes after the decimal but before the 7 don't count, and 7, 0, 6, and 2 count as significant figures.

So, the answer would be 0.007062.

Let me know if this helps :)

8 0
2 years ago
A gas sample occupies 200 mL at 760 mm Hg. What volume does the gas occupy at 400 mm Hg?
astra-53 [7]

Answer:

201.9

Explanation:

when you divided 760 with 400 yo get 19.0 the add it with 200 you get that answer

3 0
3 years ago
WILL GIVE BRAINLIEST TO PERSON WHO ANSWERS FASTEST AND CORRECT BUT ANY ANSWERS TO JUST GET POINTS WILL GET REPORTED AND DELETED
AleksandrR [38]

Answer:

B

Explanation:

When water is at the surface of the ground, it evaporates and goes back to the clouds for which causes it to rain therefore returning water back.

3 0
2 years ago
How did Neils Bohr change the model of the atom
kipiarov [429]
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6 0
2 years ago
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

5 0
3 years ago
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