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 : The mass of solute in solution is .
Solution : Given,
Molarity = 0.730 M
Volume of solution = 1.421 L
Molar mass of sodium carbonate = 105.98 g/mole
Formula used for Molarity :
where,
w = mass of solute
M = Molar mass of solute
V = volume of solution in liter
Sodium carbonate is solute and water is solvent.
Now put the given values in above formula, we get the mass of solute in solution.
By rearranging the terms, we get
Therefore, the mass of solute in solution is .
C₂H₂ + O₂ --> CO₂ + H₂O
when balanced
2C₂H₂ + 5O₂ --> 4CO₂ + 2H₂O
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:
The answer to your question is: letter c (96%)
Explanation:
Indium -113 (112-9040580 amu) ₁₁₃In
Indium-115 (114.9038780 amu) ₁₁₅In
Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In
Formula
Atomic mass = m₁(%₁) +m₂(%₂) / 100
%₁ = x I established this is an equation
%₂ = 100 - x
Substituting values
114.82 = 112.8040x + 114.9039(100-x) /100 and know we expand and simplify
114.82 = 112.8040x + 11490.39 - 114.9039x /100
11482 = 112.8040x -114.9039x +11490.39
11482 - 11490.39 = 112.8040x -114.9039x
-8.39 = -2.099x
x = 3.99
Then % of Indium-115 = 100 - 3.99 = 96