The amount of particles (any kind) in ONE mole (of any substance) is always equal to Avogadro's number: 6.02*10+23
(This is the definition of a 'Mole'. It is just a number like a 'dozen' -12- or 'gross' -144-, though much larger. There is nothing 'chemical' in this number, however)
So, here is your answer: 0.56 mole = 0.56 * 6.02*10+23 = 3.37*10+23
Pretty close to one mole, Avogadro's number, of atoms.
56 grams iron (1 mole Fe/55.85 grams)(6.022 X 10^23/1 mole Fe)
= 6.038 X 10^23 atoms of Fe ( call it 6.04 X 10^23 )
1.00274 moles
There are more atoms of sulfur in 16 grams than there are atoms of iron in 56 grams.
This is a mass stoichiometry problem. Start with the balanced equation: CaCO3 --> CaO + CO2. Do a conversion from 50g CaO to moles: 56g/1mol=50g/x, x=.9 moles. The equation is balanced as written, with all coefficients understood to be 1. So: .9 moles CaO means .9 moles CaCO3. Do another conversion from moles to grams: 100g/1mol=x/.9 moles. Solve for x to get 90 grams. (56g=molar mass of calcium oxide; 100g=molar mass of calcium carbonate.)
Well, I ran the numbers for what you have in the question, and it didn't make sense, so I'm assuming you really mean 65g of iron yields 33g iron oxide. I'm also assuming you mean iron(III) oxide, which is iron's most common valence state when it reacts with oxygen, which means the product's formula would be Fe2O3.With those things in mind, here's your balanced equation:4Fe + 3O2 --> 2Fe2O3.Step1: given 65g Fe, how many moles is that? Direct proportion between 65g and the unknown # of moles and iron's atomic mass of 56g/mol gives us 1.16 moles of iron.Step 2: molar ratio of iron to iron(III) oxide is 4-2, which simplifies to 2-1, so you will have 0.58 moles of product.Step 3: given 0.58 moles of product, set up another direct proportion with the unknown mass over given moles equal to iron(III) oxide's molar mass of 160g/mol to get a mass of 92.8g theoretical yield of product.Step 4: your actual yield was 33g, so divide that by the stoichiometric yield of 92.8g, which is 0.355, which means about a 36% actual yield, which is the answer.
56g N2= 4 moles 25 c = 298.15 J=K 750mmHG= .987 atm r=.08206 v=(nRT)/P v= (4*.08206*298.15)/.987 v= 99.2 l r= a constant.
56g
Molar mass of iron is 56g. Given mass of iron= 112g No. of moles = Given mass/Molar Mass => 112g/56g= 2 moles
This is carbon tetra chloride. there are 0.3636 moles in this mass.
There are more atoms of sulfur in 16 grams than there are atoms of iron in 56 grams.
This is a mass stoichiometry problem. Start with the balanced equation: CaCO3 --> CaO + CO2. Do a conversion from 50g CaO to moles: 56g/1mol=50g/x, x=.9 moles. The equation is balanced as written, with all coefficients understood to be 1. So: .9 moles CaO means .9 moles CaCO3. Do another conversion from moles to grams: 100g/1mol=x/.9 moles. Solve for x to get 90 grams. (56g=molar mass of calcium oxide; 100g=molar mass of calcium carbonate.)
56g=0.56kg
Well, I ran the numbers for what you have in the question, and it didn't make sense, so I'm assuming you really mean 65g of iron yields 33g iron oxide. I'm also assuming you mean iron(III) oxide, which is iron's most common valence state when it reacts with oxygen, which means the product's formula would be Fe2O3.With those things in mind, here's your balanced equation:4Fe + 3O2 --> 2Fe2O3.Step1: given 65g Fe, how many moles is that? Direct proportion between 65g and the unknown # of moles and iron's atomic mass of 56g/mol gives us 1.16 moles of iron.Step 2: molar ratio of iron to iron(III) oxide is 4-2, which simplifies to 2-1, so you will have 0.58 moles of product.Step 3: given 0.58 moles of product, set up another direct proportion with the unknown mass over given moles equal to iron(III) oxide's molar mass of 160g/mol to get a mass of 92.8g theoretical yield of product.Step 4: your actual yield was 33g, so divide that by the stoichiometric yield of 92.8g, which is 0.355, which means about a 36% actual yield, which is the answer.
56g N2= 4 moles 25 c = 298.15 J=K 750mmHG= .987 atm r=.08206 v=(nRT)/P v= (4*.08206*298.15)/.987 v= 99.2 l r= a constant.
Use the equation; mass=moles*gramformulamass or m=n*gfm m=2*55.8 = 111.6g in two moles of Iron gfm or the molecular mass of a compound can be found by adding the Relative atomic masses of each element in the compound together. For example - Carbon dioxide CO2 The formula contains 1 Carbon atom and 2 Oxygen atoms (RAMs should be found in a datasheet/book or provided in the question. RAM = relative atomic mass) RAM Carbon = 12 RAM Oxygen = 16 gfm of CO2 = 12 + 32 = 44 or 1mole of CO2 = 44g
130% of 56g = 130% * 56 = 1.3 * 56 = 72.8g
56g
In one mole of each element there is... Fe - 56g O - 16g S - 32g Th equation is Fe2(SO4)3 56g x 2 = 112g 32g x 3 = 96g 16g x 12 = 192g 112g + 96g + 192g = 400g 400g
a lollypop has 24 carbs in it