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Juliette [100K]
3 years ago
12

2) write down two examples of mixturesand state their use in daily life.​

Chemistry
1 answer:
prisoha [69]3 years ago
7 0

Answer:

<em>first </em><em>example</em><em> </em><em>is</em><em>. </em><em>Air </em><em> </em><em>it </em><em>is </em><em>mixture </em><em>of </em><em>many </em>

<em>gases </em><em> </em>

<em>it's </em><em>us </em><em> </em><em>it </em><em>keeps </em><em>us </em><em>alive </em><em> </em><em>and </em><em>many </em><em>more </em><em>uses</em>

<em>second </em><em>example</em><em> </em><em>is </em><em>water </em><em>is </em><em>also </em><em>a </em><em>mixture</em><em> </em>

<em>made </em><em>up </em><em>of </em><em>carbon </em><em>and </em><em>hydrogen</em><em> </em>

<em>and </em><em>its uses </em><em>is </em><em>to </em><em>fulfill</em><em> </em><em>our </em><em>needs </em>

<em>hope </em><em>it </em><em>helps</em>

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
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Answer:

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Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

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