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lys-0071 [83]
3 years ago
5

When should a line graph be used

Chemistry
1 answer:
olga_2 [115]3 years ago
6 0

Answer:

<em>Line graphs are used to track changes over short and long periods of time. When smaller changes exist, line graphs are better to use than bar graphs. Line graphs can also be used to compare changes over the same period of time for more than one group.</em>

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Which is the larger atom kr Or As
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I think As is larger

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Using the two models, compare the processes of nuclear fusion and nuclear fission. What do the two processes have in common?
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Answer:it’s A and B

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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. nh4cl cobr3 k2so
Illusion [34]
Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result  in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

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3 years ago
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5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?
Aleksandr [31]

Answer:

V_2=9.20L

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

\frac{V_2}{n_2}= \frac{V_1}{n_1}

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L

Best regards!

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3 years ago
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