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crimeas [40]
4 years ago
6

you are given 500 mL of a 5M stock solution of ammonium chloride but for your experiment, you only need 100 mL of a 0.65M soluti

on. Explain how to make the solution you need?
Chemistry
1 answer:
mafiozo [28]4 years ago
6 0

Answer:

13ml

Explanation:

  1. to prepare dis solution first you need a volumetric flask of 100ml calibrated .
  2. using dilution formulae u will need 13ml from the stock
  3. measure 13ml of the 5M stock of the 500ml
  4. then drop it into the 100ml calibrated flask
  5. then add water till it reach d mark and you shake thoroughly
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4 years ago
Q#1 Give a combination of four quantum numbers that could be assigned to an electron occupying a 5p orbital.
Gnesinka [82]

Answer:

For an electron in 5p orbital

5,1,-1,+1/2

For an electron in 6d orbital

6,2,-2,+1/2

Explanation:

The term quantum numbers refers to a set of values that can be used to determine the region in space where it is likely to find an electron. This region in space where there is a high probability of finding an electron is known as an atomic orbital. An atomic orbital is actually a wave function according to the Schrödinger wave equation.

There are four quantum numbers used in describing an atomic orbital: the principal quantum number (n), the orbital angular momentum quantum number also called azimuthal or subsidiary quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms).

For a 5p orbital;

n= 5, l= 1, ml= -1,0,1 ms= +1/2 or -1/2

For 6d orbital;

n= 6, l= 2, ml= -2,-1,0,1,2, ms= +1/2 or -1/2

Since we are requested to use a four quantum number description that can be assigned to an electron in these levels;

For an electron in 5p orbital

5,1,-1,+1/2

For an electron in 6d orbital

6,2,-2,+1/2

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3 years ago
What is the formula or equation for<br><br>motion,speed and energy?
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3 years ago
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4 years ago
A 200.-milliliter sample of CO2(g) is placed in a sealed, rigid cylinder with a movable piston at 296 K and 101.3 kPa. Determine
bagirrra123 [75]

Answer:

The final volume of the sample of gas V_{2} = 0.000151 m^{3}

Explanation:

Initial volume V_{1} = 200 ml = 0.0002 m^{3}

Initial temperature T_{1} = 296 K

Initial pressure P_{1} = 101.3 K pa

Final temperature T_{2} = 336 K

Final pressure P_{2} =  K pa

Relation between P , V & T is given by

P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }

Put all the values in the above equation we get

101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )

V_{2} = 0.000151 m^{3}

This is the final volume of the sample of gas.

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