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insens350 [35]
2 years ago
15

Can someone plz fill the blanks? 20 points.

Chemistry
1 answer:
lawyer [7]2 years ago
3 0

Explanation:

1. cryolite

2. cell

3. bauxite

4.

5.

6.

7.

8.

9. replaced

this is all I know. I'm sorry I could not answer all

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10 molecules SO2 to moles
cluponka [151]

Answer:

1.66 * 10^-23 moles

Explanation:

If you follow dimensional analysis, molecules to moles is 10 molecules divided by 6.022 times 10 to the 23rd and that gives you your answer

3 0
4 years ago
Why is butanone achiral?
maw [93]
Because it has no <span> stereogenic carbon centres.</span>
4 0
3 years ago
4P+ 5O2 -&gt; P4O10
Ratling [72]
If 4 moles of P is used by 5 mole of O2
then....0.489 moles will be used by 5/4 × .489 = .611 moles of O2

so .611 moles

so if 4 moles of P is burnt , 1 mole of P4O10 is produced ....so for .489 moles...... .489/4=.122 moles !
so mass will be .122× 283.89 = 34.7 grams

so first ans is .611 moles and second is 34.7 grams !

if you have any problem regarding this , just comment !!!
4 0
3 years ago
Read 2 more answers
Spectator ions when FeSO4 and PbNO32 react
sweet [91]

Sorry I thought I had the answer!

Explanation:

4 0
3 years ago
Determine the freezing point of a water solution of fructose (C6H12O6) made by dissolving 92.0 g of fructose in 202 g of water.
Naya [18.7K]

Answer:

THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C

Explanation:

To calculate the freezing point of a water solution of fructose,

1. calculate the molar mass of Fructose

( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol

2. calculate the number of moles of fructose in the solution

number of moles = mass / molar mass

n = 92 g / 180 g/mol

n = 0.511 moles.

3. calculate the molarity of the solution

molarity = moles / mass of water in kg

molarity = 0.5111 / 202 g /1000 g

molarity = 0.5111 / 0.202

molarity = 2.529 M

4. calculate the change in the freezing point of pure solvent and solution ΔTf

ΔTf = Kf * molarity of the solute

Kf = 1.86 ◦C/m for water

ΔTf = 1.86 * 2.529

ΔTf = 4.70 C

5. the freezing point is therefore

0.00 ◦C - 4.70 ◦C = -4.70 ◦C

4 0
3 years ago
Read 2 more answers
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