4 years ago

A single atom of oxygen cannot be subdivided into simpler substances by any physical or chemical means. Therefore, it is:

Chemistry

2 answers:

Zigmanuir [339]4 years ago

8 0

Qn1)d
qn2)b
qn3)a
qn4)a

Goshia [24]4 years ago

7 0

1) D. An element
2) D. Suspension
3) A. Iodine
4) A. Atoms of the same element

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If 30.943 g of a liquid occupy a space of 35.0 ml. What is the density of the liquid in g/cm3?

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3 years ago

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

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

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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4 years ago

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PLS ANSWER FAST THANK UU

andriy [413]

Answer
I think it might be B
Explanation

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3 years ago