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2H2 + O2 ---> 2H2O
1 mole of H2 = 2g
1 mole of O2 = 32g
1 mole of H2O = 18g
according to the reaction:
2*2g H2---------------32g O2
1,6g H2----------------- x g O2
x = 12,8g O2
so reaction is stechiometric
32g O2---------------- 2*18g H2O
12,8g O2 -------------- x
x = 14,4g H2O
answer: 14,4g of water vapour
hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )
where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )
where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )
where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.
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"I" symbol means the current goes through the system (imagine the 'I' being a line, like a circuit connecting [power to the device]) "O" symbol means the current does not go through the system. ( the circle is an open circuit, having no power flowing through it
There are approximately 160 grams in 1 mol of fe203 molecules therefore there would be 79/160 = 0.49375 mols of fe203 molecules in 79 grams therefore 5 atoms in total for each molecule of fe203 therefore 79/160 *5 =79/32=2.46875 mols in atoms