Which pair could be genetic clones? look at the photo

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35. How many atoms of lead are in the samp

aliya0001 [1]

11.35 g atoms of lead are in the sample.

What does density mean?

The density of an area refers to the quantity of things—which may include people, animals, plants, or objects—there are in it.
Divide the number of objects by the area's measurement to determine density.
A country's population density is calculated by dividing its total population by its area, expressed in square kilometers or miles.

We can calculate the volume of the cube by cubing the side of the cube.

Volume of cube = ( 1.000 cm)³

= 1.000 cm³

We can now determine the amount of lead in the cube by multiplying by the density. Because we know we have a 1.000 cubic centimeter and that lead has a density of 11.35 grams per cubic centimeter, we have 11.35 grams of lead.

= ( 1.000 cm³) ( 11.35 $\frac{g}{cm^{3} }$ )