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makkiz [27]
3 years ago
12

Give below are the four mixtures. Choose the correct one which can be separated by winnowing. (i) soybean and chickpeas (ii) whe

at and rice (iii)cotton and cornflakes (iv)Banana wafers and chips Which of these can be separated by the method of winnowing?
Chemistry
1 answer:
victus00 [196]3 years ago
4 0

Answer:

Banana wafers and chips

Explanation:

Winnowing is the process of separating heavier and lighter components of a mixture by blowing air through the mixture. This implies that the relative weight of the particles in the mixture determines whether they can be separated by winnowing or not.

However, chips are heavier than wafers. wafers refer to a very light snack which is easily blown away by a current of air. Hence banana chips and wafers can be separated using the method of winnowing.

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A pure substance is found to contain 53.7% fluorine and 46.3% xenon by mass. What is the empirical formula of this substance?
evablogger [386]

Answer: XF8

Explanation:

Empirical Formular shows the simplest ratio of elements in a compound.

 Xe = 46.3%             F  = 53.7%

Divide the percentage composition of each element by the atomic  mass.

Xe = 46.3/ 131.3                      F= 53.7/ 19

      = 0.353( approx)               =  2.826 (approx)

Divide through with the smallest of the answers gotten in previous step.

 Xe = 0.353 / 0.353                F = 2.826/ 0.353

       =  1                                       = 8.0

Empirical formular = XF8

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3 years ago
3. A measurement of the freezing temperature of a solution allowsyou to calculate the concentration of the solution. What else d
Harman [31]

Answer:

Osmotic pressure and boiling point elevation

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In the the osmotic pressure one can determine the molar mass of a solid by calculating the number  of  moles from the Morality formula which involves the volume of the solution.

In the boiling point elevation you can determine the number of moles of the solute in the solution by using the Molality formula.

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3 years ago
How many kL does a 9.51 ´ 109 cL sample contain?
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Answer:

9.51 × 10⁴ kL

Explanation:

Step 1: Given data

Volume of the sample (V): 9.51 × 10⁹ cL

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9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL

9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.

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