26,4 g iridium is equal to 0,137 mol.
To find the number of moles in 26.35 grams of iridium, you first need to determine the molar mass of iridium, which is approximately 192.22 g/mol. Then you can use the formula: moles = mass / molar mass. Substituting the values, moles = 26.35 g / 192.22 g/mol ≈ 0.137 moles of iridium.
To find the number of atoms in 179.0 g of iridium, first determine the molar mass of iridium, which is approximately 192.22 g/mol. Then, calculate the number of moles in 179.0 g by dividing the mass by the molar mass: ( \text{moles} = \frac{179.0 , \text{g}}{192.22 , \text{g/mol}} \approx 0.93 , \text{mol} ). Finally, multiply the number of moles by Avogadro's number (( 6.022 \times 10^{23} , \text{atoms/mol} )) to find the total number of atoms: ( 0.93 , \text{mol} \times 6.022 \times 10^{23} \approx 5.59 \times 10^{23} , \text{atoms} ).
To find the number of atoms in 179.0 g of iridium (Ir), first determine its molar mass, which is approximately 192.22 g/mol. Next, calculate the number of moles in 179.0 g by dividing the mass by the molar mass: 179.0 g / 192.22 g/mol ≈ 0.933 moles. Finally, multiply the number of moles by Avogadro's number (approximately (6.022 \times 10^{23}) atoms/mol) to find the total number of atoms: (0.933 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 5.61 \times 10^{23}) atoms.
14,84 g magnesium are equivalent to 0,61 moles.
26,4 g iridium is equal to 0,137 mol.
To find the number of moles in 26.35 grams of iridium, you first need to determine the molar mass of iridium, which is approximately 192.22 g/mol. Then you can use the formula: moles = mass / molar mass. Substituting the values, moles = 26.35 g / 192.22 g/mol ≈ 0.137 moles of iridium.
To find the number of atoms in 179.0 g of iridium, first determine the molar mass of iridium, which is approximately 192.22 g/mol. Then, calculate the number of moles in 179.0 g by dividing the mass by the molar mass: ( \text{moles} = \frac{179.0 , \text{g}}{192.22 , \text{g/mol}} \approx 0.93 , \text{mol} ). Finally, multiply the number of moles by Avogadro's number (( 6.022 \times 10^{23} , \text{atoms/mol} )) to find the total number of atoms: ( 0.93 , \text{mol} \times 6.022 \times 10^{23} \approx 5.59 \times 10^{23} , \text{atoms} ).
To find the number of atoms in 179.0 g of iridium (Ir), first determine its molar mass, which is approximately 192.22 g/mol. Next, calculate the number of moles in 179.0 g by dividing the mass by the molar mass: 179.0 g / 192.22 g/mol ≈ 0.933 moles. Finally, multiply the number of moles by Avogadro's number (approximately (6.022 \times 10^{23}) atoms/mol) to find the total number of atoms: (0.933 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 5.61 \times 10^{23}) atoms.
The formula is: number of moles = g Be/9,012.
27.4 g H2O x 1 mole/18 g = 1.52 moles
14,84 g magnesium are equivalent to 0,61 moles.
97,5 g of oxygen is equal to 5,416 moles.
978 g calcium contain 24,4 moles.
67,4 g HCl is equivalent to 1,85 moles.
29,0 g of calcium is equal to 0,723 moles.
573,28 of g of AgCI is equivalent to 4 moles.