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How many moles are in 95.0 gram of octane?
how many moles are in 95.0 gram of octane?
A piece of aluminum foil 0.550 mm thick and 1.00 cm squared is allowed to react with bromine to form aluminum bromide B How many grams of aluminum bromide form?
To calculate the grams of aluminum bromide (AlBr₃) formed from aluminum foil, we start with the volume of aluminum foil, which can be found by multiplying its thickness by its area: (0.550 , \text{mm} \times 1.00 , \text{cm}^2 = 0.0550 , \text{cm}^3). The density of aluminum is approximately 2.70 g/cm³, so the mass of aluminum is (0.0550 , \text{cm}^3 \times 2.70 , \text{g/cm}^3 = 0.1485 , \text{g}). From the balanced chemical reaction, 2 moles of Al react with 6 moles of Br₂ to produce 2 moles of AlBr₃. Thus, 0.1485 g of Al (approximately 0.0055 moles) will yield about 0.0055 moles of AlBr₃, which corresponds to (0.0055 , \text{moles} \times 266.69 , \text{g/mol} \approx 1.47 , \text{g}) of aluminum bromide.
How many moles of sodium in 6.5 gram of sodium?
The answer is 0,111 moles.
How many atoms are in 1.46 moles of aluminum fluoride?
1,46 moles of aluminum fluoride contain 35,16848.10e23 atoms.
How many moles of lithium fluoride are in an 85.2 gram sample?
85.2 gram LiF sample is equivalent to 3,28 moles.
How many moles are in 95.0 gram of octane?
how many moles are in 95.0 gram of octane?
How many moles of aluminum are in 1.99 grams of aluminum?
1,99 grams of aluminum is equal to 0,0737 moles.
A piece of aluminum foil 0.550 mm thick and 1.00 cm squared is allowed to react with bromine to form aluminum bromide B How many grams of aluminum bromide form?
To calculate the grams of aluminum bromide (AlBr₃) formed from aluminum foil, we start with the volume of aluminum foil, which can be found by multiplying its thickness by its area: (0.550 , \text{mm} \times 1.00 , \text{cm}^2 = 0.0550 , \text{cm}^3). The density of aluminum is approximately 2.70 g/cm³, so the mass of aluminum is (0.0550 , \text{cm}^3 \times 2.70 , \text{g/cm}^3 = 0.1485 , \text{g}). From the balanced chemical reaction, 2 moles of Al react with 6 moles of Br₂ to produce 2 moles of AlBr₃. Thus, 0.1485 g of Al (approximately 0.0055 moles) will yield about 0.0055 moles of AlBr₃, which corresponds to (0.0055 , \text{moles} \times 266.69 , \text{g/mol} \approx 1.47 , \text{g}) of aluminum bromide.
How many moles are there in 54 gram of water?
The answer is 3 moles.
How many moles of aluminum are in 10grams of aluminum?
10 grams aluminum (1 mole Al/26.98 grams) = 0.37 moles of aluminum ---------------------------------
How many moles of sulfur are present in 3 moles of aluminum sulfate?
There are 6 moles of sulfur present in 3 moles of aluminum sulfate, because aluminum sulfate has a 2:3 ratio of aluminum to sulfur.
How many moles are in a gram of carbon?
1 gram carbon (1 mole C/12.01 grams) = 0.08 moles carbon ===============
When 4 moles of aluminum are allowed to react with an excess of chlorine gas Cl2 how many moles of aluminum chloride are produced?
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
How many moles of sodium in 6.5 gram of sodium?
The answer is 0,111 moles.
How many brands of aluminum foil are there?
lots
How many atoms are in 1.46 moles of aluminum fluoride?
1,46 moles of aluminum fluoride contain 35,16848.10e23 atoms.
How many grams of aluminum chloride are produced when 18 grams of aluminum are reacted with an excess of hydrochloride acid?
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
