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Softa [21]
3 years ago
9

After doing a correct and careful recrystallization, you isolate your product by filtration. Which of the following will be pres

ent in the mother liquor? (select all that apply)
a. Impurities
b. Large amount of product
c. Recrystallizing solvent
d. Small amount of produc
Chemistry
1 answer:
Anna71 [15]3 years ago
3 0

Answer:

a. Impurities

c. Recrystallizing solvent

Explanation:

In this type of reaction the products are never considered totally pure, that is why as a final product it must always be taken into account that it is proportional, it will be a recrystallization solvent and other impurities with which the products were mixed.

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I believe the answer is .87g/mL. But I'm not sure so whatevs
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PLEEEASSEEEE HELPPPPP!!!!!Take some time to research a utility plant. If there is one in your area, you may even visit it.
raketka [301]

Answer:

There are four laws of thermodynamics that define fundamental physical quantities (temperature, energy, and entropy) and that characterize thermodynamic systems at thermal equilibrium.

Explanation:

6 0
3 years ago
The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

5 0
3 years ago
How many grams are there in 7.5x10^23 molecules of H2SO4
horsena [70]
Moles of H2SO4= 7.5x10^23/ 6.02x10^23 = 1.25 (3sf) moles of H2SO4

Mass of 1 mole of H2SO4= 98.1g

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6 0
3 years ago
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer
lidiya [134]

The question is incomplete, the complete question is

Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V

Answer:

only IV and V

Explanation:

If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.

On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.

This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.

7 0
3 years ago
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