It is sodium oxide Na2O.
To find the mass of sodium oxide formed, we first need to calculate the moles of sodium used, which is 0.3 moles (6.9g / 23g/mol). Since one mole of sodium reacts with one mole of oxygen to form sodium oxide, the moles of sodium oxide formed is also 0.3 moles. The molar mass of sodium oxide (Na2O) is 62g/mol. Therefore, the mass of sodium oxide formed is 18.6 grams (0.3 moles * 62g/mol).
The formula mass of sodium oxide, Na2O is 2(23.0)+16.0 = 62.0 Amount of sodium oxide = 7.94/62.0 = 0.128mol
To find the molarity of the solution, you first need to calculate the mass of ethylene glycol in the solution. Then convert it to moles using the molar mass of ethylene glycol. Finally, calculate the molarity by dividing the moles of ethylene glycol by the volume of the solution in liters.
What mass of will be produced from the given masses of both reactants? 28.0 g/P4 / 123.9 g/P4 = 0.2260 moles/ P4 0.2260 moles /P4 * 4 moles PCl5/1 mole/P4=0.904 moles/PCl5 54.0 g/Cl2 / 70.9 g/Cl2 = .7616 moles/Cl2 .7616 moles / Cl2 *4 moles PCl5/10 moles Cl2=.30 This is about limiting reagents You need to use 2P + 5Cl2 --> 2PCl5 [or P4 + 10Cl2 --> 4PCl5 if you prefer] that tells you that 2P = 62g needs 5Cl2 = 355g Cl2 to react 69.3g
To find the percent yield, first calculate the theoretical yield of ammonia based on the given amounts of nitrogen and hydrogen. Then compare the actual yield (62g) with the theoretical yield to calculate the percent yield using the formula: (actual yield/theoretical yield) x 100%. The percent yield would be the actual mass of ammonia produced (62g) divided by the theoretical yield of ammonia.
To find the mass of sodium oxide formed, we first need to calculate the moles of sodium used, which is 0.3 moles (6.9g / 23g/mol). Since one mole of sodium reacts with one mole of oxygen to form sodium oxide, the moles of sodium oxide formed is also 0.3 moles. The molar mass of sodium oxide (Na2O) is 62g/mol. Therefore, the mass of sodium oxide formed is 18.6 grams (0.3 moles * 62g/mol).
The formula mass of sodium oxide, Na2O is 2(23.0)+16.0 = 62.0 Amount of sodium oxide = 7.94/62.0 = 0.128mol
To find the amount of oxygen used, we need to consider the difference in mass between sodium and sodium oxide. The mass increase is 16g (62g - 46g) which corresponds to the amount of oxygen used from the air. Therefore, 16g of oxygen from the air were used.
Assuming you mean 80 cubic centimeters, the density would be 62/80 or 0.775gms/cc
Balanced equation: 2Na(s) + O2(g) ---> Na2O(s)moles Na = 115 g x 1 mol/23 g = 5 molesmoles Na2O possible = 1/2 x 5 moles = 2.5 moles (b/c the mole ratio of Na2O:Na is 1:2)mass Na2O possible = 2.5 moles x 62 g/mole = 155 grams (assuming O2 is not limiting)
density = mass / volume = 62g / 21.9 cm³ ≈ 2.8 g/cm³
usual 62g
If you are referring to grams, 62g is approximately equal to 2.186985638 ounces.
A 20oz bottle of Squirt contains 62g of sugar.
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mw : 134 /2= 67
To find the molarity of the solution, you first need to calculate the mass of ethylene glycol in the solution. Then convert it to moles using the molar mass of ethylene glycol. Finally, calculate the molarity by dividing the moles of ethylene glycol by the volume of the solution in liters.