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.
2 significant figures.
56g
40 g / 56 g X 100%
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.)
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
56g=0.56kg
Density = Mass/Volume = 0.622... grams per ml
Assuming the volume is in cubic cm and not cm(!), Density = Mass/Volume = 56g/28cm3 = 2 grams per cm3
130% of 56g = 130% * 56 = 1.3 * 56 = 72.8g
Density = Mass/Volume => V = M/D So Vol = 25.8/2.56 = 10.1 units (approx) of the volume in which the density was expressed.
2 significant figures.
Molar mass of iron is 56g. Given mass of iron= 112g No. of moles = Given mass/Molar Mass => 112g/56g= 2 moles
56g
A standard tennis ball weighs about 56g-58g.
here is a code for membership mee 3d5 56g ht6 8ty
2 oz (That's 56g in grown-up measurement).
1 cup = 56 grams