Consider a piece of gold jewelry that weighs 9.85 g and has a volume of 0.675 cm3 . the jewelry contains only gold and silver, which have densities of 19.3 g/cm3 and 10.5 g/cm3, respectively. if the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains, calculate the percentage of gold (by mass) in the jewelry.

61.5096 %

Explanation:

Let the amount of gold in the jewelry = x g

Let the amount of silver in the jewelry = y g

Total mass of jewelry = 9.85 g

So,

x + y = 9.85 g    .................................1

Density = Mass / Volume

Given that:

Density of gold = 19.3 g/cm³

Thus, volume is:

Volume = Mass / Density = x / 19..3 cm³

Given that:

Density of silver = 10.5 g/cm³

Thus, volume is:

Volume = Mass / Density = y / 10.5 cm³

The total volume of the jewelry = 0.675 cm³

So,

Or,

10.5 x + 19.3 y = 136.78875  ................2

Solving 1 and 2, we get that:

x = 6.0587 g

y = 3.7913 g

So, mass % of gold:

Mass % = (6.0587 / 9.85)×100 = 61.5096 %

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

when atmosphere interacts with biosphere they focuses on on the sources and sinks of green house gases (ghgs) air pollutants, particular matters, water and energy between the earth's surface and the atmosphere. the biosphere affects the climate. the atmosphere and biosphere are in constant interaction with each other, and all living things rely on gases in the atmosphere for life. organisms such as animals take in oxygen gas form the atmosphere to use in every cell of their body for smooth functioning of daily routine exercises.