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2 years ago

How is the structure of a hammer related to its function

1 answer:

7 0

The long handle gives leverage so you can pound nails in and the round end is meant for pounding nails and the tongs end is for pulling them out

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If charcoal and sugar were the the main impurities present in crude acetanilide would recrystallisation from water be a good met

siniylev [52]

Yes, recrystallization from water would be a viable technique of purification if crude acetanilide contained sugar and charcoal. Due to their physical insolubility, charcoal found in crude acetanilide and water will form a heterogeneous combination. Charcoal will separate during crystallization because it is insoluble in water. Sugar and the unreconstituted crude acetanilide make up the remaining portion of the combination. Crude acetanilide will crystallize in this combination and separate from the sugar and water solution. Filtration will be used to capture the crude acetanilide that has solidified. Sugar and water make a soluble solution, thus they are kept separate. This is how the charcoal and sugar combination is purified and its constituent parts are separated from crude acetanilide using the recrystallization process from water.

4 0

2 years ago

Please help I’m torn between A and B. I’ve asked many and people say A, but I asked my science teacher and she said B. And also

Vladimir79 [104]

Just choose one of them. Getting one question wrong isn't the end of the world.

I suggest water.

8 0

3 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

1.2 g of sucrose is added to water with the final volume of 75 ml. What is the concentration of sucrose in %w/v?

Firlakuza [10]

Answer:

16 g/L

Explanation:

Given that:-

Mass of sucrose added = 1.2 g

Volume of water = 75 mL = 0.075 L ( 1mL = 0.001 L )

%w/v is defined as the mass of the solution in 1 L of the solution.

So,