Answer:
Explanation:
*Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7. On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.
*pH at the equivalence point =7
we know that pH + pOH = 14
Hence pOH= 14-7=7
pOH= -log(OH-)
The concentration of OH-= 10-pH= 1X10-7 M
One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.
Protons:
- Have a mass
- Positively charged
- Found inside the nucleus of an atom
Electrons:
- Have a mass. (9.10938188×10−31 kilograms), though this can sometimes be considered negligible due to how small that actually is. Barely factored into atomic mass
- Negatively charged
- Found outside the nucleus in the electron shell
Neutrons:
- Have a mass
- Neutral (no charge)
- Found inside the nucleus of an atom
Atom A:
- 1 proton
- 0 Neutrons
- 1 electron
- Atomic mass of 1
- Atomic number of 1
Atom B:
- 8 Protons
- 10 Neutrons
- 8 electrons
- Atomic mass of 18
- Atomic number of 8
Atomic mass includes the number of protons and neutrons in the nucleus. Atomic number is the number of protons, as this is what defines what type of element the atom is.
Answer:
The molecule has a bent geometry
Explanation:
Let us look again at the principles of VSEPR theory. The shape of a molecule depends on the number of electron pairs that surround the valence shell of the central atom in the molecule.
Lone pairs distort the molecular geometry away from what is expected on the basis of VSEPR theory.
The molecule described in the question has the form AEX2. Two substituents and one lone pair form three electron domains around the central atom. The expected geometry is trigonal planar but the observed molecular geometry is bent because of the lone pairs present.
Answer:
Explanation:
The electron in the lowest energy state will be found in 1 s energy level.
set of 4 possible quantum numbers
Principal quantum no : n = 1 ,
Azimuthal quantum no l = 0
Magnetic quantum no m = 0
Spin quantum no s = + 1/2
set of other quantum nos
Principal quantum no : n = 1 ,
Azimuthal quantum no l = 0
Magnetic quantum no m = 0
Spin quantum no s = - 1/2