When the substance is said to be an oxide of iron, that means the compound contains elements of Fe and O. Suppose you have 100 g of the sample.
Mass of Fe = 72.36 g
Mass of O = 100 - 72.36 = 27.64 g
The molar mass of Fe is 55.8 g/mol while the molar mass of O is 16 g/mol.
Moles of Fe = 72.36/55.8 = 1.29677
Moles of O = 27.64/16 = 1.7275
Now, divide both amounts by the lower number, which is 1.29677.
Fe: 1.29677/1.29677 = 1
O = 1.7275/1.29677 = 1.333
The answer must be whole numbers. So, let's try multiplying the least number to O that would make it a whole number. That would be 3.
Fe: 1*3 = 3
O: 1.333*3 = 4
<em>Thus, the empirical formula is Fe₃O₄.</em>
Answer:
5.06atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (Litres)
V2 = final volume (Litres)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 1.34 atm
P2 = ?
V1 = 5.48 L
V2 = 1.32 L
T1 = 61 °C = 61 + 273 = 334K
T2 = 31 °C = 31 + 273 = 304K
Using P1V1/T1 = P2V2/T2
1.34 × 5.48/334 = P2 × 1.32/304
7.34/334 = 1.32P2/304
Cross multiply
334 × 1.32P2 = 304 × 7.34
440.88P2 = 2231.36
P2 = 2231.36/440.88
P2 = 5.06
The final pressure is 5.06atm
I think it is eutrophication
Star is rats backwards if that helps