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Free_Kalibri [48]
2 years ago
14

Give an example where handpicking is used for separation. (1)​

Chemistry
1 answer:
dimulka [17.4K]2 years ago
5 0

Answer:

If you want to separate black grapes from the mixture of black and green grapes, then you will simply pick black grapes using your hands from the mixture. In this way you are actually using handpicking separation method.

Explanation:

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What is the correct answer and why?
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7 0
3 years ago
A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.
Andru [333]

Answer:  The empirical formula and the molecular formula of the organic compound is CHO and C_4H_4O_4 respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2 = 21.71 g

Mass of H_2O= 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, =\frac{12}{44}\times 21.71=5.921g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, =\frac{2}{18}\times 5.926=0.658g of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53  g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles

Mass of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.493}{0.493}=1

For H =\frac{0.658}{0.493}=1

For O=\frac{0.658}{0.493}=1

The ratio of C : H: O =  1: 1: 1

Hence the empirical formula is CHO.

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4

Thus molecular formula = n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4

6 0
2 years ago
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