8.9mt
One cubic foot of common red brick weigh 120 pounds. For fire clay bricks, the weight is 150 pounds per cubic foot.
The weight of 1 cu. ft. of aluminum is approximately 169 pounds. It is also equivalent to 76,500 grams or 76.5 kilograms. The chemical symbol for aluminum is Al.
Nothing by itself. The cubic foot must be full of some kind of matter to have weight. Obviously, a cubic foot of feathers will weigh far, far less than a cubic foot of lead.
1 cu ft = 0.037037 cu yeard
A 2.18 moles of Cu has a weight of 138.53 g. This is obtained by multiplying the 2.18 by the weight of 1 mole of Cu.
1900000000000000
I assume you mean 4.6 X 10^25 atoms. This is the conversion. 4.6 X 10^25 atoms Cu ( 1mol Cu/ 6.022 X 10^23 )( 63.55 grams Cu/ 1 mol Cu ) = 4854.37 grams Cu
8.9mt
The molar mass of copper is its atomic weight on the periodic table in g/mol, and is 63.5g/mol We know that one mole of copper contains 6.022×10^23 atoms of copper . First convert given mass to moles, and moles to atoms. = 61.0 g Cu × (1 mol Cu / 63.5 g per mol) ×6.022 × 10 ^23 atom cu / 1 mol Cu) = 5.78× 10^23. atoms. 61 g Cu 5.7 ×10^23 atoms of Cu.
62.4 pounds
62.30 pounds per cu ft
63.546 g/mol
The mole in chemistry is also called the chemist's dozen and is defined as the amount of material containing 6.0221421 X10^23 particles(This number is called Avogadro's number) The value of mole is the number of particles in excactly 12 grams of c-12, so, if you have 12grams of c-12 , you will have 6.022x10^23 carbon atoms ,which is also a mol of C. For any other element a mol of that element is the Atomic Mass expressed as grams. 0.0265 g C find mol of C plan gC -> mol C 1 mol / 12.01 g C ( relationship; 1 mol C = 12.01 g C ) 0.0265 g C x 1 mol C / 12.01 g C = 2.21 x 10 ^-3 mol C to find atoms change to mol then times 6.022X10^23 3.10g Cu find Cu atoms plan g -> mol cu -> atoms Cu (3.10 g cu )x (1 mol Cu /63.55 g Cu ) ( 6.022 x 10^23 / 1 mol cu = 2.94 x 10^22 Cu atoms
10 pounds
50 lb.
Cu + 2NaOH ---> Cu(OH)2 + 2NaSince we have a finite amount of two reactants we must first determine which is the limiting reactant. For Cu: n= m/M = 0.500 g/63.53 g/mol = 7.87E-3 mol For NaOH: n = CV = (3.0 mol/L)(0.0300 L) = 9.00E-2 mol Copper is by far the limiting reactant therefore we will use its amount to find the maximum Cu(OH)2 that can be yielded from the reaction. Since the above reaction scheme indicates that Cu and Cu(OH)2 are in a 1 to 1 ratio, their molar amounts are the same ie. at the end of the reaction there will be 7.87E-3 mol of Cu(OH)2 To find the mass that corresponds to this molar amount we multiply by the molar mass of Cu(OH)2 (97.56 g/mol): m=nM = (7.87E-3 mol)(97.56 g/mol) = 0.7678 g Therefore 0.500 g of Cu will yield 0.768 g (3 sig fig) of Cu(OH)2.