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stiv31 [10]
3 years ago
8

How does the electron configuration of elements within the same group compare? (A.) They all have their valence electrons in the

same energy level. ( B) They all have their valence electrons in the same type of subshell. (C) They all have their valence electrons in a p suborbital. ( D) They all have their valence electrons in the 3rd energy level.
Chemistry
1 answer:
lesya [120]3 years ago
5 0

Answer:

B.They all have their valence electrons in the same type of subshell

Explanation:

  • The electron configurations of elements in the same group (column) of the periodic table have them in the same type of subshell.
  • But the subshells may be of different shells. Thus , the energies of them need not be the same.
  • For example , The Alkalai Metals are found in the first column of the periodic table Group IA. This set of elements all have valence electrons in only the 's' orbital and because they are in the first column they all have s^{1} configuration. i.e,
  1. H 1s^1
  2. Li 1s^2 2s^1
  3. Na 1s^2 2s^2 2p^6 3s^1
  4. K 1s^2 2s^2 2p^6 3s^2 3p^64s^1
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(5.625 + 8.15) x 2.34 + 3.2
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The answer is 35.4335

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An atom that normally has __________ electrons in its outer (valence) shell tends not to form chemical bonds with other atoms.
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8 valence shell electrons.
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Assume that the maximum number of ATPs is produced (38). At pH 7, and in the presence of excess Mg2 , how much of the energy in
Lubov Fominskaja [6]

In one mole of glucose 38 ATP energy is stored this accounts for only 40 per-cent of the total energy in glucose.

Explanation:

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8 0
3 years ago
Please help!!
Olegator [25]
Molarity =  Moles/Liter

Use the molecular atomic mass of NaCl to convert from grams to moles.
Molecular mass of NaCl is the sum of its atomic masses. Look at the periodic table to find these. Na is 23 g/mol and Cl is 35.5 g/mol ,
so NaCl = 23 + 35.5 = 58.5 g/mol

multiply to cancel out grams
76 g NaCl * (1mol / 58.5 g NaCl) = 1.3 mol NaCl

over 1 Liter is just 1.3 M NaCl
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3 years ago
A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

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