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blagie [28]
3 years ago
11

Suppose your teacher asks you to separate a mixture of sand and water. Which of the following methods would you use: distillatio

n, filtration, or evaporation? Explain your answer.
Chemistry
2 answers:
weqwewe [10]3 years ago
8 0
To separate, filtration would work best. Sand particles and water molecules differ drastically in size. So when filtered, the sand would not flow out of the device as the water would. This will effectively separate the water and sand with the smallest amount of material lost.

Hope this helps :)
larisa86 [58]3 years ago
7 0
Depends, does the teacher just care about the water and sand being separated or does the teacher want you to separate them and still have both substances in their original form. If the teacher just cares about the separation I'd use evaporation. It was still water and sand the only thing is the water changed forms and became a gas. If the teacher wants both substances in their original state then I'd use filtration. You can separate the sand from the water and still have them both.
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How is the preodictable arranged by atomic mass, coloms , and rows
bearhunter [10]

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Explanation:

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5 0
2 years ago
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Based on the equation 3 Cu (s) + 8 HNO3 (aq) 3 Cu(NO3)2 (aq) + 2NO (g) + 4H2O (g) how many grams of Cu would be needed to react
eimsori [14]

Answer:

Mass = 381.28 g

Explanation:

Given data:

Number of moles of HNO₃ = 16 mol

Mass of Cu needed to react with 16 mol of HNO₃ = ?

Solution:

Chemical equation:

3Cu + 8HNO₃    →     3Cu(NO₃)₂ + 4H₂O + 2NO

Now we will compare the moles of Cu with HNO₃ from balance chemical equation.

                     HNO₃         :          Cu

                        8              :         3

                       16             :       3/8×16 = 6

Mass of Cu needed:

Mass = number of moles × molar mass

Mass = 6 mol × 63.546 g/mol

Mass = 381.28 g

4 0
3 years ago
For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution?
gogolik [260]
By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
             = -㏒ 10^-6
             = 6
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3 years ago
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8 0
3 years ago
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ratelena [41]

Answer:

B

Explanation:

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8 0
1 year ago
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