85 moles Li x 6.02x10^23 atoms/mole Li = 5.17x10^25 atoms of Li
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.
For this you need the atomic (molecular) mass of NaNO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaNO3=85.0 grams60.1 grams NaNO3 / (85.0 grams) = .707 moles NaNO3
4.25 g NaNO3 x 1 mole NaNO3/85 g = 0.05 moles0.05 moles/0.1 L = 0.5 moles/L = 0.5 M
There are 85 inches in 85 inches.
To calculate the molality of the solution, we first need to determine the number of moles of NaNO3. The molar mass of NaNO3 is approximately 85 g/mol, so 20.0 g of NaNO3 corresponds to about 0.235 moles (20.0 g / 85 g/mol). Molality (m) is defined as the number of moles of solute per kilogram of solvent. With 750.0 g of water (0.750 kg), the molality is 0.235 moles / 0.750 kg = 0.313 m.
85g of oxygen = 85/16 moles of O atoms = 5.3125 moles There are 2 moles of oxygen atoms per mole of carbon dioxide, so we have 2.65625 moles of carbon dioxide. This is 2.65625 x 44 g = 116.875 g.
To find the number of moles, you need to divide the given mass (85 grams) by the molar mass of AgNO3 (169.87 g/mol). 85 grams of AgNO3 represents 0.500 moles.
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.
To find the number of moles, first calculate the molar mass of sodium nitrate (NaNO3), which is 85 grams/mol. Then, divide the given mass (2.85 grams) by the molar mass to obtain the number of moles present, which is approximately 0.0335 moles.
To find the number of moles of H3O in the solution, you can use the formula pH -logH3O. First, calculate the concentration of H3O ions using the pH value: pH -logH3O 3.0 -logH3O H3O 10(-3.0) 1.0 x 10(-3) M Next, calculate the number of moles of H3O in the solution using the concentration and volume: moles concentration x volume moles 1.0 x 10(-3) mol/L x 85 L moles 8.5 x 10(-2) moles Therefore, there are 8.5 x 10(-2) moles of H3O present in the 85 L solution with a pH of 3.0.
For this you need the atomic (molecular) mass of NaNO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaNO3=85.0 grams60.1 grams NaNO3 / (85.0 grams) = .707 moles NaNO3
There are 85 inches in 85 inches.
4.25 g NaNO3 x 1 mole NaNO3/85 g = 0.05 moles0.05 moles/0.1 L = 0.5 moles/L = 0.5 M
It depends on 85% of what!
7400/85
An 85-key piano has 85 keys.
85 inches is 215.9cm