How many atoms are in 766.3 grams of Li?

1 answer:

34kurt4 years ago

6 0

1 mol of any substance is made of 6.022 x 10²³ units, these units could be atoms making up elements or molecules making up a compound.

In this case. 1 mol of Li is made of 6.022 x 10²³ atoms of Li

The molar mass of Li is 6.94 g/mol

Therefore mass of 1 mol of Li is 6.94 g

In 6.94 g of Li there are - 6.022 x 10²³ atoms

Then in 766.3 g of Li there are - 6.022 x 10²³ atoms / 6.94 g x 766.3 g = 665 x 10²³

There are 6.65 x 10^25 atoms of Li

You might be interested in

Energy released from atomic reactions is called?

erma4kov [3.2K]

The answer is nuclear energy

7 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: