How many moles are in 3.90 x1030 atoms of Zn?

1 answer:

7 0

Explanation:
The problem basically wants you to find a way to convert between the number of atoms present in the sample and the number of moles they are equivalent to.
To convert between atoms and moles we use something called Avogadro's constant, which basically acts as the definition of a mole.
More specifically, in order to have one mole of an element you need
6.022 x 10^23 atoms of that element. You can thus use this number as a conversion factor to take you from atoms to moles or vice versa.
In your case, you will have

3.90 x 10^ 26 atoms Zn x 1 mole Zn ( Avogrado’s constant) / 6.022 x 10^23 atoms Zn

= 6.5 x 10^8 is the answer

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Some forms of vitamin D, C₂₈H₄₄O, can be found in red meat. How many total atoms are in 5 molecules of C₂₈H₄₄O

MaRussiya [10]

Answer:

3.01 × 10^24 atoms of vitamin D

Explanation:

The number of atoms, molecules or ions present in a substance is given by the Avogadro's number which is 6.02 × 10^23.

Hence;

1 molecule of vitamin D contains 6.02 ×10^23 atoms

5 molecules of vitamin D contains 5 × 6.02 ×10^23/1

= 3.01 × 10^24 atoms of vitamin D

7 0

3 years ago

Find the percent composition of each element in KMnO4

VARVARA [1.3K]

The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!

6 0

3 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$