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dlinn [17]
3 years ago
11

Which of the following is not a valid set of four quantum numbers

Chemistry
1 answer:
AfilCa [17]3 years ago
3 0

Answer:

The incorrect quantum numbers are 1,1,0,+\frac{1}{2}

Explanation:

1,1,0,+\frac{1}{2}

n=1,l=1,m_{1}=0,m_{s}=+\frac{1}{2}

It is not a valid set of quantum number, "l" value cannot be taken from "n" value.

2,0,0,+\frac{1}{2}

n=2,l=0,m_{1}=0,m_{s}=+\frac{1}{2}

This set of quantum number represents second energy level.

1,0,0,+\frac{1}{2}

n=1,l=0,m_{1}=0,m_{s}=+\frac{1}{2}

This set of quantum number represents first energy level.

2,1,0,-\frac{1}{2}

n=2,l=1,m_{1}=0,m_{s}=-\frac{1}{2}

This set of quantum number represents 2p energy level.

3,0,0,-\frac{1}{2}

n=3,l=0,m_{1}=0,m_{s}=-\frac{1}{2}

This set of quantum number represents third energy level.

Therefore, The incorrect quantum numbers are 1,1,0,+\frac{1}{2}

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A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
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Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

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V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

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How much heat energy is required to raise the temperature of 0.360 kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of co
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MThe  heat  energy  required  to  raise  the  temperature   of  0.36Kg   of  copper   from   22 c   to  60  c  is  calculate  using  the  following  formula

MC delta T
m(mass)=  0.360kg  in  grams  =  0.360  x1000 = 360 g
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heat  energy is therefore=  360g   x0.0920 cal/g.c  x 37  c=  1225.44  cal

5 0
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