The molar mass of copper (Cu) is approximately 63.55 g/mol. To calculate the weight of 2.18 mol of copper, you would multiply the number of moles by the molar mass: 2.18 mol x 63.55 g/mol ≈ 138.49 g. Therefore, 2.18 mol of copper weighs approximately 138.49 grams.
The balanced chemical equation for the reaction between copper (Cu) and nitric acid (HNO3) is Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O. From the equation, it can be seen that 1 mol of Cu reacts with 4 mol of HNO3. Therefore, to react with 2.0 mol of HNO3, 0.5 mol of Cu is needed. The molar mass of Cu is approximately 63.5 g/mol, so 0.5 mol of Cu would be equivalent to 31.75 grams.
The molecular weight of copper (Cu) is approximately 63.55 g/mol.
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.
To calculate the number of Cu atoms in 85 mol, you need to use Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol. So, for 85 mol of Cu, the number of Cu atoms would be 85 mol x 6.022 x 10^23 atoms/mol = 5.12 x 10^25 Cu atoms.
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
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.
To find the number of atoms in 1.2 grams of copper, you need to first determine the molar mass of copper (Cu). The molar mass of copper is 63.55 g/mol. Next, calculate the number of moles in 1.2 grams of copper (1.2 g / 63.55 g/mol = 0.0189 mol). Finally, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms: 0.0189 mol x 6.022 x 10^23 atoms/mol = approximately 1.14 x 10^22 atoms.
The molar mass of Cu is 63.55 g/mol and of O is 16 g/mol. The moles of Cu used is 127g / 63.55 g/mol = 2 mol, and the moles of O used is 32g / 16 g/mol = 2 mol. Since the mole ratio of Cu to O in CuO is 1:1, then 2 mol of Cu will react with 2 mol of O to form 2 mol of CuO. The mass of 2 mol of CuO is 159.6g.
2.6 * 6.022 * 10^23 atoms Explanation- 1 mol of anything has 6.022 * 10^23 units. So 2.6 mol has 2.6 times this number. Also Cu is monoatomic Note: * means multiplied; ^ means raised to the power
To find the weight of 4.6 x 10^25 atoms of copper, you can start by calculating the molar mass of copper, which is approximately 63.55 g/mol. Next, convert the number of atoms to moles by dividing by Avogadro's number (6.022 x 10^23). Finally, multiply the number of moles by the molar mass to find the weight in grams.
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