There you go , This is my work i have did on that subject area
<em>V = 151 mL = 151 cm³</em>
<em>d = 0,789 g/mL = 0,789 g/cm³</em>
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d = m/V
m = d×V
m = 0,789×151
<u>m = 119,139g</u>
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.
The last photo because its molecular formula is C3H4.
Answer: 150 kPa
Explanation:
Given that,
Original volume of gas V1 = 30L
Original pressure of gas P1 = 105 kPa
New pressure of gas P2 = ?
New volume of gas V2 = 21L
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
105 kPa x 30L = P2 x 21L
3150 kPa L = P2 x 21L
P2 = 3150 kPa L / 21 L
P2 = 150 kPa
Thus, 150 kPa of pressure is required to compress the gas