The answer is 2,511.1023 molecules.
13 L of FeO2 can be produced from 50.0 L of O2 at STP.
0.361 moles O2
The answer is 8,5379.10e23 molecules.
The answer is o,5 moles MgO.
The mass is 3,358 kg.
13 L of FeO2 can be produced from 50.0 L of O2 at STP.
22,4 L is the molar volume. 1 mol of oxygen has 32 g and and 6,022140857.10e23 molecules.
At STP1mol O2 = 22.4L1mol O2 = 6.022 x 1023 molecules O222.4L O2 = 6.022 x 1023 molecules O2Convert liters O2 to moles O23.36L O2 X (6.022 x 1023 molecules O2/22.4L O2) = 9.03 x 1022 molecules O2
0.361 moles O2
The answer is 8,5379.10e23 molecules.
You should set up a balanced equation for the combustion of CH4 first: CH4 + 2 O2 --> CO2 + 2 H2O Now you see that for every 1 mole of CH4 you have 2 moles of O2. Therefore if you have 2.67 L of CH4, you'll have 5.34 L of O2. You can also assume it's at STP just to check, since it all cancels out in the end: 2.67L CH4 (1mol CH4/22.4L)(2mol O2/ 1 mol CH4)(22.4L/1mol O2)= 5.34L O2.
at stp 1 mole of a gas contains 22.4 litres. 9.1/22.4= .40625 moles o2. 1 mole of a gas contains 6.022E23 molecules so .40625 moles x 6.022E23 = 2.4464325E23 molecules, but you have to multiply by two due to it being diatomic, so answer x 2 = 4.892875E23 molecules
55 grams O2 × (1 mol O2 ÷ 32 g O2) × (2 mol H2O ÷ 1 mol O2) × (22.4 L ÷ 1 mol H2O) = 77 L H2O(g)In the liquid form, it works out to just under 62 milliliters.
2 if it has the 2.2 L engine, 3 if it has the 4.3 L engine.
The answer is o,5 moles MgO.
How much water?? ml or L
The mass is 3,358 kg.