1.656 times 10 superscript 23
0,15 mol gold is 29,545 g.
To find the number of gold atoms in 45.0 grams of gold, first determine the molar mass of gold (Au), which is approximately 197 g/mol. Next, calculate the number of moles in 45.0 grams by dividing the mass by the molar mass: ( \frac{45.0 \text{ g}}{197 \text{ g/mol}} \approx 0.228 \text{ mol} ). Finally, multiply the number of moles by Avogadro's number ((6.022 \times 10^{23} \text{ atoms/mol})): ( 0.228 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 1.37 \times 10^{23} \text{ atoms} ). Thus, there are approximately (1.37 \times 10^{23}) gold atoms in 45.0 grams of gold.
1 mol = 6,022 140 857(79).10e23 atoms.0,25 mol is 1,50553521425.10e23 atoms.
To vaporize gold, we need to consider its molar enthalpy of vaporization. The molar enthalpy of vaporization of gold is approximately 334 kJ/mol. Since the molar mass of gold is about 197 g/mol, vaporizing 2 kg (2000 g) of gold requires: ( q = \frac{2000 , \text{g}}{197 , \text{g/mol}} \times 334 , \text{kJ/mol} \approx 3,385 , \text{kJ}. ) Thus, approximately 3,385 kJ of energy is required to vaporize 2 kg of gold.
To find the number of moles of gold in a 1.00 kg piece, we first convert the mass to grams: 1.00 kg = 1000 g. The molar mass of gold (Au) is approximately 197 g/mol. Using the formula for moles (moles = mass/molar mass), we calculate: 1000 g / 197 g/mol ≈ 5.08 moles of gold.
0,15 mol gold is 29,545 g.
To find the mass of a 32 mol gold coin, you would first need to know the molar mass of gold, which is approximately 197 g/mol. Then, you can calculate the mass by multiplying the number of moles by the molar mass: 32 mol * 197 g/mol = 6304 g.
---- Golds molecular weight is 196.96 g/mol Thus, 1g of gold is (1g /196.96 g/mol) = 0.005 mol ---- 1 mol is 6.022 × 1023 atoms (Avogadro's Number) Thus, 0.005 mol is (0.005 mol x 6.022 × 1023 atoms/mol) = 3.057 x 1021 atoms ---- Therefore 1g of gold has APPROXIMATELY 3.057 x 1021 atoms
To determine the number of gold atoms in the sample, you can use the molar mass of gold (197 g/mol) to first find moles, then convert moles to atoms using Avogadro's number (6.022 x 10^23 atoms/mol). First, find moles: 5.00 x 10^-3 g ÷ 197 g/mol = 2.54 x 10^-5 mol. Then, convert moles to atoms: 2.54 x 10^-5 mol x 6.022 x 10^23 atoms/mol = 1.53 x 10^19 atoms.
To calculate the amount of gold in moles, we need to first determine the mass of gold in the medal. From the information provided, 71.5% of the medal's mass is gold. Therefore, the mass of gold in the medal is 0.715 * 115 grams. Next, we convert this mass into moles using the molar mass of gold, which is approximately 197 grams/mol. The amount of gold that could be extracted from the medal in mol is (0.715 * 115) /197.
There are approximately 0.013 moles of gold in a troy ounce. This calculation is based on the molar mass of gold, which is 196.97 g/mol.
To find the number of moles of gold, first calculate the molar mass of gold, which is 197 g/mol. Then, divide the given mass of gold by the molar mass to find the number of moles. In this case, 3.60x10-5g ÷ 197 g/mol ≈ 1.83x10-7 moles of gold.
A millimeter is 1/1000 of a meter, which is 1/1000 of a kilometer. 27.5 millimeters = .0275 meters. .0275 meters = .0000275 kilometers. 27.5 millimeters = .0000275 kilometers.
There are about 2.56 x 10^22 atoms in 10 grams of gold. This calculation is based on the atomic mass of gold (197 g/mol) and Avogadro's number (6.022 x 10^23 atoms/mol).
The leading zero has no effect, so they are the same.
To calculate the number of atoms in 0.02 g of gold (Au), you first need to determine the number of moles of gold in 0.02 g using the molar mass of gold (196.97 g/mol). Then, you use Avogadro's number (6.022 x 10^23 mol^-1) to convert moles to atoms. The calculation would be 0.02 g Au / 196.97 g/mol Au × 6.022 x 10^23 atoms/mol.
(3.28 g of chloroform/.0275moles of chloro)= 119 g of chloroform. Just take the weights in grams and divide it by number of moles. Hope it helps!