7.5g
Explanation:
Given parameters:
Mass of Fe₂O₃ = 25g
Unknown:
Mass of oxygen that can be removed = ?
Solution:
Find the ratio of the oxygen atoms to that of the compound using the molar mass. Then multiply with the given mass.
Mass of oxygen = }[/tex] x given mass
Mass of oxygen = x 25 = 7.5g
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Percentage composition brainly.com/question/8654167
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Answer:
Explanation:
From the question, one can work out which states of matter to assign to which species. The trick with organic equations of this nature is to try to balance everything but oxygen first. Make sure you balance oxygen last because it is the easiest to balance.
Answer:
Explanation:
To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.
1. Convert Atoms to Moles
To convert atoms to grams, Avogadro's number must be used.
This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.
Multiply the given number of atoms by Avogadro's number.
Flip the fraction so the atoms of iodine will cancel out.
Multiply so the problem condenses into 1 fraction.
2. Convert Moles to Grams
Now we must use the molar mass of iodine, which is found on the Periodic Table.
Use this mass as a fraction.
Multiply this fraction by the number of moles found above.
Multiply. The moles of iodine will cancel.
The 1 as a denominator is insignificant.
3. Round
The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.
The 8 in the hundredth place tells us to round the 0 up to a 1.
There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.
Answer: 4.1 g of barium precipitated.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number of particles.
To calculate the moles, we use the equation:
Given : moles of barium = 0.030
Molar mass of barium = 137 g/mol
x= 4.1 g
Thus there are 4.1 g of barium that precipitated.