0.545 g of aluminum burns completely in oxygen, producing 1.030 g of aluminum oxide. find the empirical formula of the oxide​

1.030-0.545=0.485 grams of oxygen in compound.

Convert grams of Al and grams of O into moles:

0.545g Al x 1 mol Al/26.98g Al = 0.0202 mol Al

0.485g O x 1 mol O/16.00g O = 0.0303 mol O

Next we divide the moles of both of them by the lowest number to get simplest ratio

0.0202/0.0202 = 1 Al

0.0303/0.0202 = 1.5 O

Next we multiply all numbers until we get a whole number for each one. In this case 2 works:

Al 1 x 2 : O 1.5 x 2

and we are left with Al2O3

Explanation:

Hope this helps :D

mass Al = 0.545 g

\text{mass O = 0.485 g}mass O = 0.485 g

Required:

empirical formula

Solution:

Step 1: Calculate the number of moles of each element

n \: \text{Al = 0.545 g} \times \frac{\text{1 mol}}{\text{26.98 g}} = \text{0.0202 mol}nAl = 0.545 g×

26.98 g

1 mol

=0.0202 mol

n \: \text{O = 0.485 g} \times \frac{\text{1 mol}}{\text{16.00 g}} = \text{0.0303 mol}nO = 0.485 g×

16.00 g

1 mol

=0.0303 mol

Step 2: Represent an empirical formula

empirical \: formula = \text{Al}_{x}\text{O}_{y}empiricalformula=Al

x

O

y

Step 3: Divide the number of moles of each element by the least number of moles

x = \frac{\text{0.0202 mol}}{\text{0.0202 mol}} = 1 × 2 = 2x=

0.0202 mol

0.0202 mol

=1×2=2

y = \frac{\text{0.0303 mol}}{\text{0.0202 mol}} = 1.5 × 2 = 3y=

0.0202 mol

0.0303 mol

=1.5×2=3

Step 4: Write the empirical formula

\blue{empirical \: formula = \text{Al}_{2}\text{O}_{3}}empiricalformula=Al

2

O

3

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a, b, c - sides of a triangle

a + b > c

a + c > b

b + c > a

we have a = 5m and b = 11m. substitute:

5 + 11 > c

16 > c → c < 16

5 + c > 11         subtract 5 from both sides

c > 6

11 + c > 5         subtract 11 from both sides

c > -6

c < 16 and c > 6 and c > -6

thereofre your answer is 6m < c < 16m