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algol13
3 years ago
10

Which of the following statements about electronic configurations of atoms and mononuclear ions, and atomic orbital quantum numb

ers, are true or false.
a. The electronic configuration of the hydride anion is 1s2.
b. The valence electronic configuration of strontium is 4d2.
c. For a given value of l the number of possible values of ml is 2l + 1.
d. Cu+ has the same electronic configuration as Ni.
e. The magnetic quantum number is never larger than the principle quantum number (for a given orbital).
Chemistry
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

a. The electronic configuration of the hydride anion is 1s2. TRUE

b. The valence electronic configuration of strontium is 4d2.  FALSE

c. For a given value of l the number of possible values of ml is 2l + 1.  TRUE

d. Cu+ has the same electronic configuration as Ni.  TRUE

e. The magnetic quantum number is never larger than the principle quantum number (for a given orbital). TRUE

Explanation:

a. The electronic configuration of the hydride anion is 1s² is true since the hydriden anion is the hdrogen atom which has gained an electron and we will add that electron to the 1s¹ configuration of H.

b. The valence electronic configuration of strontium is 4d2 is false since Sr is an element of period 5 , therefore its valece electrons are in in period five and it has 2 electrons because Sr belongs to group 2.

c. For a given value of l the number of possible values of ml is 2l+1 is true since this number gives the magnetic orientation for the sublevel. Thus for s there is only one orientation, then ml = 2 (0 ) +1 . Por p with  l equal to 1 we have three possible orientations : 2(1) + 1 =3. The d and f sublevels have 10 and 14 orientations.

d. Cu⁺ has the the same electronic configuration as Ni is true since Cu, atomic number 29, has one more electron than its neighbor Ni with an atomic number of 28. If we remove one electron from copper we are gong to have the same 28 electrons niquel has in its neutral state.

e. The magnetic quantum number is never larger than the principal quantum  number for a given orbital is true since l, the magnetic quantum number can have values up to n-1, the principal quantum number.

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Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

Step 1: Data given

Volume of 1.00 M Ba(NO3)2 = 1.00 L

Temperature = 25.0 °C

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enthalpy change is – 26 kJ per mol BaSO4

The specific heat of water is 4.18 J/g ·˚C

the density of water is 1.00 g/mL

Step 2: The balanced equation

Ba(NO3)2(aq) + Na2SO4(aq) → 2NaNO3(aq) + BaSO4(s)

Step 3: Calculate the total volume

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Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

Step 5: Calculate moles BaSO4 formed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

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⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

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