50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
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Here we have to get the spin of the other electron present in a orbital which already have an electron which has clockwise spin.
The electron will have anti-clockwise notation.
We know from the Pauli exclusion principle, no two electrons in an atom can have all the four quantum numbers i.e. principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s) same. The importance of the principle also restrict the possible number of electrons may be present in a particular orbital.
Let assume for an 1s orbital the possible values of four quantum numbers are n = 1, l = 0, m = 0 and s = .
The exclusion principle at once tells us that there may be only two unique sets of these quantum numbers:
1, 0, 0, + and 1, 0, 0, -
.
Thus if one electron in an orbital has clockwise spin the other electron will must be have anti-clockwise spin.
Answer: kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:
→
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide: -1002.82 kJ/mol
Calcium chloride: -795.8 kJ/mol
Water: -285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
So, the standard enthalpy of formation of HCl is -173.72 kJ/mol