Assuming you mean 138.08 grams, you set up a direct proportion between the molar mass of NO2 and your given mass. NO2 comes in at 46g/mol (14+16+16), and your proportion should be 46/1=138.08/x. Solve for x to get approx. 3 moles.
The reaction to form nitrogen dioxide using nitric oxide is; 2NO(g) + O2(g) -> 2NO2(g) As the stoichiometry between the substances are 1:1, 1.35 moles of nitrogen monoxide is needed.
3
0.98 mol NO2 (6.022 X 10^23/1 mole NO2) = 5.90 X 10^23 atoms of nitrous oxide
The answer is 5,213.10 ex.23.
Ga is Gallium and NO2 is the nitrite anion. Thus, Ga(NO2)3 is gallium nitrite.
Balanced equation. 3NO2 + H2O -> 2HNO3 + NO 8.44 moles NO2 (1 mole NO/3 moles NO2) = 2.81 moles NO formed
4.651024 molecules of NO2 equals 7,721 moles.
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will get half the number of moles of NO2. So, you will have 0.2 moles of nitric acid.
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will end up with half the moles of nitrogen dioxide (NO2)...so you will have 0.2 moles.
Molecular mass of nitrogen dioxide, NO2 = 14.0+2(16) = 46.0Amount of NO2 = 25.6/46.0 = 0.557molThere are 0.557 moles of NO2 in a 25.6g sample.
Molar mass NO2 = 46.0 g/mole1.18 g NO2 x 1 mol NO2/46.0 g = 0.0257 moles NO2 (to 3 significant figures)
4.67
4.50 moles NO2 X (46 grams NO2) / (1 mole NO2) x (4 moles NO2) / (7 moles O2) x (1 mole O2) / (32 gm O2) = 3.70 grams O2
0.41
10 grams NO2 (1 mole NO2/46.01 grams) = 0.217 moles nitrogen dioxide ======================
Determine the molar mass of NO2 using the subscripts in the formula and the atomic weights in grams from the periodic table. 1 mole NO2 = (1 x 14.0067g N) + (2 x 15.9994g O) = 46.0055g NO2 Calculate the moles NO2 by dividing the given mass by the molar mass. 25.5g NO2 x (1mol NO2/46.0055g NO2) = 0.554mol NO2
138g