Moles=given wt./molecular weight
1.25=given wt./70
given wt.=70*1.25=87.5 gm
We first calculate the molar mass of P4O10, which is 284 g/mol. Then we convert the given mass of P4O10 (142 g) to moles using its molar mass. There is a 1:4 molar ratio between P4O10 and O2 in the reaction, so we multiply the moles of P4O10 by 4 to find the moles of O2 needed.
If a 10 pence coin weighs 6.5 grams, then approximately 15 coins would weigh around 100 grams.
To find the number of grams of helium in the balloon, you need to use the ideal gas law equation. First, convert the temperature to Kelvin by adding 273 (303 K). Then, convert the pressure to atm (0.966 atm). Now, use the ideal gas law formula: PV = nRT, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature in Kelvin. Solve for n (moles), which equals 0.08 moles of helium. Finally, calculate the mass using the molar mass of helium (4 g/mol) multiplied by the number of moles (0.08 mol), which equals 0.32 grams of helium.
one ritz cracker equals 3 grams therefore you will need 5 ritz crackers which will equal 15g but it's close enough :P
To calculate the volume of the gas, you would need the ideal gas law equation: PV = nRT. First, convert grams of argon to moles using the molar mass of argon. Then, plug the values of pressure (P), temperature (T), and gas constant (R) into the equation with the calculated number of moles (n) to solve for volume (V).
Quite a few! 392 grams phosphorous (1 mole P/30.97 grams) = 12.7 moles phosphorous ===================
2.3 grams P (1mole P/30.97 grams) = 0.07427 moles Phosphorous 0.07427 moles P * 6.022 X 10^23 = 4.5 X 10^22 atoms of P in 2.3 grams P 4.5 X 10^22/6.022 X10^23 = 0.07427 moles of atoms in 2.3 grams of phosphorous
There are 2.04 vaginas in a mole so techinically speaking there 5 7 moles in your moms vagina! which mean thats 7 moles will die from lack of food! which they will then weigh .6grams each. So 7 times 2.04 vaginas is 14.28 moles. Then you take 14.28 moles times .6 grams which equals sperm! :P
Say you have 100 grams of the compound. If so, then you would have 3.1 grams H, 31.5 grams P, and 65.4 grams O. Convert these to moles: 3.1 grams H ÷ 1.01 grams/mol = 3.07 moles H 31.5 grams P ÷ 31.0 grams/mol = 1.02 moles P 65.4 grams O ÷ 16.0 grams/mol = 4.09 moles O Divide each of these by the smallest number, which is 1.02: 3.07 ÷ 1.02 = 3 1.02 ÷ 1.02 = 1 4.09 ÷ 1.02 = 4 These are your ratios. The empirical formula is then H3PO4
1000
From every two atoms of P, one molecule of P2O5 is formed. The relative formula mass of two atoms of P (phosphorous) is 62, whereas for one molecule of P2O5 it is 142. The mass of P2O5 formed is therefore 142/62 times the mass of P we started with. This comes out to 286.3g (to one decimal place).
This can be simplified to -10p = -150 p = -150/-10 p = 15
Molar mass of P2O5 is 142gmol-1. You need 17.2 moles of p.
There are 5.10 grams of Phosphorus, which has a molar mass of approximately 30.97 g/mol. This would be approximately 0.165 moles of Phosphorus. There are approximately 6.022 x 10^23 atoms in a mole, so 5.10 grams of Phosphorus would contain approximately 9.92 x 10^22 atoms.
There are 24 mol of P atoms in 96 mol of P4O10. This is because each P4O10 molecule contains 4 P atoms, so you need to divide the total moles of P4O10 by 4 to find the moles of P atoms.
The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used: n=moles m=mass (grams) M= molecular weight (from periodic table) 2Al + 3S --> Al2S3 nAl= m/M = 9/13 = 0.692307 moles of Al nS= m/M = 8/16 = 0.5 moles of S Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio) This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3 nAl2S3 = m/M Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74 0.16666 moles Al2S3 = m/74 m = 74 x 0.16666 = 12.33 grams of Al2S3 Therefore, 12.33 grams of Al2S3 is formed in this reaction ( hopefully this is right :P )
Use the ideal gas law, PV=nRT. P= pressure V= volume n= number of moles R= gas law constant T= temperature If you have P, V, R, T then you can solve for "n" to find the number of moles. There are a number of ways and variations that you can go about finding the number of moles, but all would involve the ideal gas law or a similar formula.