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How many moles are in 125g of Zn?
The equivalent of 125 g zinc is 1, 91 moles.
How many mole in 3.65 g Zn?
Using the molar mass of zinc (65.38 g/mol), you can calculate the number of moles in 3.65 g of Zn by dividing the given mass by the molar mass. So, 3.65 g of Zn is equivalent to approximately 0.056 moles of Zn.
What is the amount in moles of 3.80g Zn?
To find the amount in moles of 3.80g of zinc (Zn), you need to divide the mass by the molar mass of Zn, which is approximately 65.38 g/mol. 3.80g Zn / 65.38 g/mol ≈ 0.058 moles Zn
What is the amount in moles of 3.45 g Zn?
Calculated by dividing mass (in grams) by atomic mass of Zn: 3.45(g) / 65.38(g/mole) = 0.0528 mole
How many moles are in 3.60 g Zn?
To find the number of moles in 3.60 g Zn, you need to divide the mass by the molar mass of Zn. The molar mass of Zn is approximately 65.38 g/mol. So, 3.60 g Zn divided by 65.38 g/mol is equal to approximately 0.055 moles of Zn.
How many moles are in 125g of Zn?
The equivalent of 125 g zinc is 1, 91 moles.
How many mole in 3.65 g Zn?
Using the molar mass of zinc (65.38 g/mol), you can calculate the number of moles in 3.65 g of Zn by dividing the given mass by the molar mass. So, 3.65 g of Zn is equivalent to approximately 0.056 moles of Zn.
What is the amount in moles of 3.80g Zn?
To find the amount in moles of 3.80g of zinc (Zn), you need to divide the mass by the molar mass of Zn, which is approximately 65.38 g/mol. 3.80g Zn / 65.38 g/mol ≈ 0.058 moles Zn
What is the amount in moles of 3.45 g Zn?
Calculated by dividing mass (in grams) by atomic mass of Zn: 3.45(g) / 65.38(g/mole) = 0.0528 mole
What is the mass in grams of 0.040 moles Zn?
0,040 moles Zn equal 2,614 g.
How many grams of Zn are in 0.356 moles of zinc?
0,356 moles of zinc contain 23,27 g.
How many moles are in 3.60 g Zn?
To find the number of moles in 3.60 g Zn, you need to divide the mass by the molar mass of Zn. The molar mass of Zn is approximately 65.38 g/mol. So, 3.60 g Zn divided by 65.38 g/mol is equal to approximately 0.055 moles of Zn.
How many grams of ZnCl2 are formed from150 g of Zn and 73 g of HCl?
To determine the amount of ZnCl2 formed, we first need to find the limiting reactant. Zn is the limiting reactant in this case. The balanced chemical equation is: Zn + 2HCl -> ZnCl2 + H2. Using the given mass of Zn, calculate the moles of Zn, then use the mole ratio from the balanced equation to find the moles of ZnCl2 formed. Finally, convert the moles of ZnCl2 to grams.
How many zinc atoms are present in 20.0 g Zn?
Molar mass of Zinc = 65.4 g mol-1 No. of moles of 20.0 g of Zinc = 20.0 / 65.4 = 0.305810397 mol No. of atoms = 0.305810397 L (where L is the Avogadro constant) = 1.8416 x 1023
How many moles are in 22.5 grams of zinc?
For this you need the atomic mass of Zn. Then take the mass in grams and divide it by the atomic number (multiplied by one mole for units to cancel) to find number of moles. Zinc's atomic mass is 65.4 grams.22.5 g Zn / (65.4 grams) = .344 moles Zn
How many moles of ZnCl2 will be produced from 28g of Zn assuming CuCl2 is available in excess?
Ok, so we have... Zn + CuCl2 -> ZnCL2 + Cu And we are given that we have 28g of Zn (Zinc). But the question calls for moles of ZnCl2. So, the first step is to convert grams to moles of Zn. We will need to find the AMU (Atomic Mass Units) that a Zn atom's mass. A Zn atom has a mass of 63.546 AMU. Now, we take the given mass and divide it by the mass of the atom to get the moles of reactant we have. 28g/63.546AMU≈ .4406257moles If your teacher/ professor cares about sig figs, then it's .44 because you only have two sig figs from the given from the given mass of Zn (28g). Now, the easy part. What is the ratio of Zn atoms on the reactant side to Zn atoms on the product side? 1:1 And since we have excess CuCl2 Nothing else limits the reaction. So, .44 (or how ever far your teacher wants you to round) moles of Zn reactants in a ideal environment (you will learn later that no reaction is 100% and that there are reactants always left over and how to solve for that) produces .44 moles of ZnCl2.
How many milliliters of 4.50 M HCl(aq) are required to react with 9.65 g of Zn(s)?
First, calculate the moles of Zn using its molar mass. Then, use the mole ratio from the balanced chemical equation between Zn and HCl to find the moles of HCl needed. Finally, use the molarity of HCl to calculate the volume in milliliters using the formula: Volume (mL) = moles / molarity.
