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The answer is 1,72 moles.

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Q: How many moles of 02 gas are in 38.5L of O2 gas?
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How many grams in 15 moles of 02?

15 moles of 02 equal 480 g.


How many grams of water are made from complete reaction of 2.2 moles of oxygen gas with hydrogen gas?

The balanced chemical equation for the reaction between oxygen and hydrogen is2H2 + 02 -> 2H2OThus 2.2 moles of oxygen reacts with 4.4 moles of hydrogen to form 4.4 moles of steam (water in gaseous state).The mass of H2O obtained is thus 4.4 x 18.0 = 79.2g.


How many oxygen atoms does 1 mole of 02 gas have?

1 mole of 02 gas has 12,044 281 714.1023 atoms.


How many moles of potassium carbonate are in 10.0mL of a 2.0 M solution?

Ten milliliters is a hundredth of a liter. So in a two molar solution, you would have .02 moles in 10 ml.


What is the volume of 0.05mol of neon gas stp?

The volume of 5.0 moles of 02 at STP is 100 litres.


How many moles of oxygen gas there in 135 L of oxygen at STP?

Since we know that one mole of any gas at STP is equal to 22.4 L we can multiply 135L by the following conversion: 1 mole/22.4L. When you set up the problem it looks like this: (135L)x 1 mole/22.4L =6.03 moles of oxygen gas The liters cancel out and you are left with moles as your units. Remember, if you have liters and want moles, divide by 22.4 liters; if you have moles and you want liters you multiply by 22.4 liters.


What is 02 gas?

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Gas engine has two, diesel has zero.


How many moles of NaCl are contained in 100o ml of a 0.20 M solution?

Molarity = moles of solute/Liters of solutionMoles of solute = Liters of solution * Molarity ( 100 mL = 0.1 Liters )Moles of NaCl = 0.1 Liters * 0.20 M NaCl= 0.02 moles NaCl============


How many moles of C3H8 are needed to react with 0.567 mol of 02?

You have to burn C3H8 in O2. You get 3CO2 plus 4H2O. So to burn one mole of C3H8, you need 5 moles of O2. That means you need one fifth of C3H8 as compared to O2. So you need 0.567/5 = 0.1134 moles of C3H8. Hence the answer.


Is 02 a greenhouse gas?

no, but Ozone is


How many grams of 02(g) are needed to completely burn 23.7 g of C38H(g)?

The balanced equation is C3H8 + 5O2 ---> 3CO2 + 4H2O moles C3H8 = 23.7 g x 1 mol/44 g = 0.539 moles moles O2 needed = 5 x 0.539 moles = 2.695 moles O2 (it takes 5 moles O2 per mole C3H8) grams O2 needed = 2.695 moles x 32 g/mole = 86.2 grams O2 needed (3 sig figs)