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Is there 143g mol in 6.80g of al2o3?
To determine if there are 143 g/mol in 6.80 g of Al₂O₃, we first need to calculate the molar mass of Al₂O₃. The molar mass of Al₂O₃ is approximately 102 g/mol (with aluminum at about 27 g/mol and oxygen at about 16 g/mol). To find the number of moles in 6.80 g of Al₂O₃, we divide the mass by the molar mass: 6.80 g ÷ 102 g/mol ≈ 0.067 moles. Thus, there are no 143 g/mol in 6.80 g of Al₂O₃; the molar mass is actually around 102 g/mol.
How many moles of Al(CN)3 are in 235 g of the compound?
To find the number of moles of Al(CN)3 in 235 g, first calculate the molar mass of Al(CN)3. Aluminum (Al) has a molar mass of approximately 26.98 g/mol, and cyanide (CN) has a molar mass of about 26.02 g/mol, so the molar mass of Al(CN)3 is approximately 26.98 + (3 × 26.02) = 105.04 g/mol. Finally, divide the mass of the compound by its molar mass: 235 g ÷ 105.04 g/mol ≈ 2.24 moles of Al(CN)3.
What is the percent yield of this reaction if 750.0 g of aluminum oxide produced 256.734 g of aluminum?
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)
How many grams of aluminum would be required to produce 550 g of Fe?
To produce 550 g of Fe, you would need the same amount of Al based on the balanced chemical equation of the reaction between Al and Fe. The molar ratio between Al and Fe is 2:3. Therefore, you would need (2/3) * 550 g of Al, which is approximately 367 g.
How many grams of Fe can be produced when 10.0 g of Al is reacted with an unlimited supply of Fe2o3?
To determine how many grams of Fe can be produced from 10.0 g of Al reacting with Fe2O3, we first need the balanced chemical equation for the reaction: [ 4Al + 3Fe2O3 \rightarrow 4Fe + 6Al2O3 ] From the equation, 4 moles of Al produce 4 moles of Fe. The molar mass of Al is approximately 27 g/mol, so 10.0 g of Al corresponds to about 0.37 moles of Al. This produces an equivalent of 0.37 moles of Fe. Given that the molar mass of Fe is about 56 g/mol, the mass of Fe produced is approximately (0.37 , \text{moles} \times 56 , \text{g/mol} \approx 20.72 , \text{g}).
Which equation correctly represents the first ionization of aluminum?
The first ionization of aluminum is Al(g) -> Al+(g) + e-
When was Al G. Wright born?
Al G. Wright was born in 1916.
What is the word AL--G----AL?
allegorical, perhaps
How many grams of aluminum are in 0.471 moles of aluminum please show your work?
The molar weight of aluminum (Al) is 26.98 g/mol. grams of Al = 26.98 x 0.471 = 12.71 g The molar weight of aluminum (Al) is 26.98 g/mol. grams of Al = 26.98 x 0.471 = 12.71 g The molar weight of aluminum (Al) is 26.98 g/mol. grams of Al = 26.98 x 0.471 = 12.71 g
How many moles of Al(CN)3 are in 229 g of the compound?
The molar mass of Al(CN)3 is 105.0337 g/mol. To determine the moles Al(CN)3 in 229 g Al(CN)3, multiply the given mass by 1 mol/105.0337 g.229 g Al(CN)3 x 1 mol Al(CN)3/105.0337 g Al(CN)3=2.18 mol Ag(NO)3, rounded to 3 significant figures
Which one has more no of atoms 10 g of Al or 10 g of Fe?
10 g of Fe has more atoms because iron has a lower atomic mass than aluminum. This means that there are more atoms in 10 g of Fe compared to 10 g of Al.
Is there 143g mol in 6.80g of al2o3?
To determine if there are 143 g/mol in 6.80 g of Al₂O₃, we first need to calculate the molar mass of Al₂O₃. The molar mass of Al₂O₃ is approximately 102 g/mol (with aluminum at about 27 g/mol and oxygen at about 16 g/mol). To find the number of moles in 6.80 g of Al₂O₃, we divide the mass by the molar mass: 6.80 g ÷ 102 g/mol ≈ 0.067 moles. Thus, there are no 143 g/mol in 6.80 g of Al₂O₃; the molar mass is actually around 102 g/mol.
Calculate the number of grams of Al in 371 g of Al2O3?
To calculate the number of grams of Al in 371 g of Al2O3, you first need to determine the molar mass of Al2O3 (102 g/mol). Then, calculate the molar mass of Al (27 g/mol). From the chemical formula Al2O3, you can see that there are 2 moles of Al for every 1 mole of Al2O3. Therefore, by using these ratios, you can determine that there are 162 g of Al in 371 g of Al2O3.
How do you calculate the number of grams of Al in 286 g of Al2O3 .?
To find the number of grams of Al, first calculate the molar mass of Al2O3 (2Al + 3O). Then, find the molar mass of Al. Divide the molar mass of Al by the molar mass of Al2O3 and multiply by 286 g to get the grams of Al in 286 g of Al2O3.
How many moles of Al(CN)3 are in 235 g of the compound?
To find the number of moles of Al(CN)3 in 235 g, first calculate the molar mass of Al(CN)3. Aluminum (Al) has a molar mass of approximately 26.98 g/mol, and cyanide (CN) has a molar mass of about 26.02 g/mol, so the molar mass of Al(CN)3 is approximately 26.98 + (3 × 26.02) = 105.04 g/mol. Finally, divide the mass of the compound by its molar mass: 235 g ÷ 105.04 g/mol ≈ 2.24 moles of Al(CN)3.
What has the author Al G Field written?
Al. G. Field has written: 'Watch yourself go by' -- subject- s -: Accessible book
What is the percent yield of this reaction if 750.0 g of aluminum oxide produced 256.734 g of aluminum?
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)
