-14 J/mol
84 J/6 moles = 14 J/mole = ∆H
-14 J/Mol
84 J/6 moles = 14 J/mole = ∆H
H2O is a covalent compound predominantly, but it spontaneously ionizes to produce about 10-7 moles per liter each of hydrogen and hydroxide ions at 25C.
3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O is the equation if it is dilute nitric acid. In concentrated nitric acid the equation is different. So 3 moles of copper produce 2 moles of NO. Therefore it requires 6 moles of copper to produce 4 moles of NO.
-14 J/mol
84 J/6 moles = 14 J/mole = ∆H
-14 J/Mol
84 J/6 moles = 14 J/mole = ∆H
H2O is a covalent compound predominantly, but it spontaneously ionizes to produce about 10-7 moles per liter each of hydrogen and hydroxide ions at 25C.
N2 + 3H2 -----> 2NH3 so 3 moles of hydrogen produce 2 moles of ammonia. Therefore 12.0 moles of hydrogen will produce 8 moles of ammonia.
12 J/3 moles = 4 J/mole. Thus, H of reaction in kJ/mole = 0.004 kJ/mole
The answer is 97,66 moles.
Approx 3.29 moles.
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
The formula means, among other things, that there are 7 atoms of oxygen in each mole of the compound. Therefore, in 4.00 moles of the compound, there are 28.00 moles of oxygen atoms. Elemental oxygen usually is diatomic, so that there would be the equivalent of 14 moles of diatomic elemental oxygen.
3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O is the equation if it is dilute nitric acid. In concentrated nitric acid the equation is different. So 3 moles of copper produce 2 moles of NO. Therefore it requires 6 moles of copper to produce 4 moles of NO.