Find the Density at STP BeAuTi

$B e A u T i$

Step 1

To find the mass of $1$ mole of $B e A u T i$ look up the atomic mass of each element and multiply it by the number of atoms contained in each element in the molecule.

mass of $1$ mole of $B e A u T i$ $=$ $(mass of Be)$ + $(mass of Au)$ + $(mass of Ti)$

Step 2

Fill in the atomic masses from the periodic table.

mass of $1$ mole of $B e A u T i$ $=$ $(9.01000022)$ + $(196.97000122)$ + $(47.86999893)$

Step 3

Simplify the result.

Remove parentheses.

mass of $1$ mole of $B e A u T i = 9.01000022 + 196.97000122 + 47.86999893 g$

Add $9.01000022$ and $196.97000122$ .

mass of $1$ mole of $B e A u T i = 205.98000144 + 47.86999893 g$

Add $205.98000144$ and $47.86999893$ .

mass of $1$ mole of $B e A u T i = 253.85000038 g$

Step 4

One mole of any gas occupies $22.4 L$ at STP (standard temperature and pressure).

$V = 22.4 L$

Step 5

Density is mass over volume.

$\frac{253.85000038 g}{22.4 L}$

Step 6

Factor $253.85000038$ out of $253.85000038 g$ .

$\frac{253.85000038 (g)}{22.4 L}$

Step 7

Factor $22.4$ out of $22.4 L$ .

$\frac{253.85000038 (g)}{22.4 (L)}$

Step 8

Separate fractions.

$\frac{253.85000038}{22.4} \cdot \frac{g}{L}$

Step 9

Divide $253.85000038$ by $22.4$ .

$11.3325893 \frac{g}{L}$

Step 10

Combine $11.3325893$ and $\frac{g}{L}$ .

$\frac{11.3325893 g}{L}$

Find the Density at STP BeAuTi

Related Questions

Support LCMGCF.com by disabling your adblocker.