32.7 g x 1 mol/22.99 g = 1.422 mol x 6.022x10^23 molecules/1mol = 8.565 x 10^23 molecules(also called atoms)
To find the number of moles of Na in 42 grams, we can use the molar mass of Na, which is approximately 23 grams/mol. First, calculate the number of moles by dividing the given mass by the molar mass: 42 grams / 23 grams/mol = 1.83 moles of Na.
Balanced equation. 4Na + O2 -> 2Na2O 10 moles Na (2 moles Na2O/4 moles Na) = 5.0 moles Na2O produced
To find the number of moles of Na in 15 g of NaCl, you first need to calculate the molar mass of NaCl, which is 58.44 g/mol. Since Na makes up 39.34% of the molar mass of NaCl, you can calculate the moles of Na as (15 g / 58.44 g/mol) * 0.3934 = 0.255 moles of Na.
0.188 moles Na (6.022 X 10^23/1 mole Na) = 1.13 X 10^23 atoms of sodium
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 2.5 moles Na+ and 2.5 moles Cl-, totaling 5 moles of ions altogether.
The answer is 0,25 moles.
0.013089701
To find the number of moles of Na in 42 grams, we can use the molar mass of Na, which is approximately 23 grams/mol. First, calculate the number of moles by dividing the given mass by the molar mass: 42 grams / 23 grams/mol = 1.83 moles of Na.
0.21705 moles. There are 22.9898 grams sodium in 1 mole.
Using the equation 2Na + 2HCl -> 2NaCl + H2, we can see that 2 moles of Na react with 2 moles of HCl to produce 1 mole of H2. Calculate moles of Na: 25g Na * 1 mol Na / 23g Na = 1.09 mol Na. Since the reaction is with excess HCl, the number of moles of H2 produced will be half the number of moles of Na, so it will be 0.545 moles of H2.
Balanced equation. 4Na + O2 -> 2Na2O 10 moles Na (2 moles Na2O/4 moles Na) = 5.0 moles Na2O produced
0.125 moles x 6.02x10^23 atoms/mole = 7.53x10^22 atoms
2 moles of NaCl, of course. Cl would definitely limit in this one to one reaction and you would have 19998 moles Na in excess.
To find the number of moles of Na in 15 g of NaCl, you first need to calculate the molar mass of NaCl, which is 58.44 g/mol. Since Na makes up 39.34% of the molar mass of NaCl, you can calculate the moles of Na as (15 g / 58.44 g/mol) * 0.3934 = 0.255 moles of Na.
It depends on what you are reacting the sodium with to generate hydrogen gas. The question is incomplete and cannot be answered as it is written
To calculate the concentration of nitrate ions, we first need to find the molar mass of Mg(NO3)2, which is approximately 148.33 g/mol. Next, we calculate the number of moles of Mg(NO3)2 in 42g, which is 42g/148.33 g/mol = 0.283 moles. Since there are 2 nitrate ions per Mg(NO3)2, the number of moles of nitrate ions is 0.283 moles x 2 = 0.566 moles. Finally, we divide the moles of nitrate ions by the volume of solution in liters (0.25 L) to get the concentration: 0.566 moles / 0.25 L = 2.26 M.
42g = 1.48oz