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lesantik [10]
3 years ago
8

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M ac

etone 5 mL 1.0 M HCl 5 mL 0.0050 M I2 10 mL H2O While keeping the total volume at 25 mL and keeping the concentration of H ion and I2 as in the original mixture, how could you double the molarity of the acetone in the reaction mixture
Chemistry
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

Explanation:

pls give me point so i can ask questions

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A 5.32 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.12 g fluorine, what
bulgar [2K]

Answer:

The mass of KF in the mixture is 2.77 gms.

Explanation:

Given;

Total weight of mixture (LiF+KF)=5.32gms

Let, mass of KF in the mixture = x gms

⟹ mass of LiF in mixture =(5.32-x)gms.

We know that :

Atomic weight of F=19gms.

Atomic weight of Li =7gms.

Atomic weight of K = 39 gms.

moles=mass/(molecular weight)

Thus, moles of KF=x/58

and moles of LiF = (5.97-x)/26LiF=(5.97−x)/26

Thus,

moles of F in KF=moles of KF=x/58 ---(1)

moles of F in LiF =moles of LiF= (5.32-x)/26---(2)

From (1) & (2),

Total moles of Fluorine

=(x/58)+((5.32-x)/26)

Hence,

total weight of Fluorine in sample = moles*Atomic weight

=((x/58)+((5.32-x)/26))*19gms.

=3.12 gms.---(given)

Now, solving the equation for x,

26x +(5.32*58)-58x

=3.12*58*26/19

22x=308.56-247.62

x=60.94/22

=2.77 gms. (Answer)

Thus, the mass of KF in the mixture is 2.77 gms.

8 0
3 years ago
A certain compound is made up of one phosphorus (P) atom, three chlorine (Cl) atoms, and one oxygen (O) atom. What is the chemic
user100 [1]
<span> POCl3 is the correct way to write the chemical formula for this compound</span>
5 0
3 years ago
A reaction was performed, and the dichloromethane solvent was dried by adding magnesium sulfate drying agent. When the reaction
Degger [83]
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Sksk
7 0
3 years ago
Carbon dioxide is produced by the human body through
blsea [12.9K]

Answer:

Respiration

I hope this helps!

8 0
3 years ago
For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.
Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0
3 years ago
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