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PtichkaEL [24]
3 years ago
8

Which element has the same number of valence electrons as hydrogen (H)? A. helium (He). . B. oxygen (O). . C. nitrogen (N). . D.

lithium (Li).
Chemistry
2 answers:
velikii [3]3 years ago
7 0

Answer : The correct option is (D) lithium (Li).

Explanation :

Hydrogen is the element which belong to period 1. The atomic number of hydrogen is, 1

The electronic configuration of hydrogen is, 1s^1

The number of valence electrons present in hydrogen is, 1.

(A) Helium is the element which belong to period 1. The atomic number of helium is, 2

The electronic configuration of helium is, 1s^2

The number of valence electrons present in helium is, 2.

(B) Oxygen is the element which belong to period 2. The atomic number of oxygen is, 8

The electronic configuration of oxygen is, 1s^22s^22p^4

The number of valence electrons present in oxygen is, 6.

(C) Nitrogen is the element which belong to period 2. The atomic number of nitrogen is, 7

The electronic configuration of nitrogen is, 1s^22s^22p^3

The number of valence electrons present in nitrogen is, 5.

(D) Lithium is the element which belong to period 2. The atomic number of lithium is, 3

The electronic configuration of lithium is, 1s^22s^1

The number of valence electrons present in lithium is, 1.

From this we conclude that the lithium element has the same number of valence electrons as hydrogen.

Hence, the correct option is (D) lithium (Li).

deff fn [24]3 years ago
4 0
Li is in group 1 so it has same number of valence electrons as hydrogen.
hope it helps.
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s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

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3 years ago
Which solute produces the highest boiling point in a 0.15 m aqueous solution?
saul85 [17]
In order to determine the increase in boiling point of a solvent due to the presence of a solute, we use the formula:

ΔT = Kb * m * i

Here, Kb is a property of the solvent, so remains constant regardless of the solute. Moreover, because the concentration m has been fixed, this will also not be considered. In order to determine which solute will have the greatest effect, we must check i, the van't Hoff factor.

Simply stated, i is the number of ions that a substance produces when dissolved. Therefore, the solute producing the most ions will be the one causing the greatest change in boiling point temperature.
6 0
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13. Fill in the following table
Mandarinka [93]

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3 years ago
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san4es73 [151]

The mass of Calcium required to complete this reaction is 4.008 g.

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  • In several disciplines, including chemistry, mechanics, and fluid dynamics, the idea of mass conservation is widely applied.

In the given reaction mass of product after completion of reaction is 13.614 g that means total mass of constituents before reaction should also be 13.614.

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mass of Ca + mass of O₂ + mass of S = mass of CaSO4

Ca + 6.400 g + 3.206 g = 13.614 g

mass of Ca = 13.614 - 9.606 = 4.008 g

Therefore, by law of conservation of mass 4.008 g of Ca is required for the completion of the reaction.

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A nucleus with four protons has total positive charge
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It would have a charge of 4+
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