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2 years ago

Which of these elements is a noble gas? A. hydrogen (H) B. barium (Ba) C. nitrogen (N) D. krypton (Kr) E. bromine (Br)

Chemistry

2 answers:

Citrus2011 [14]2 years ago

7 0

Answer: The correct answer is Option D.

Explanation:

Noble gases are defined as the gases which have stable electronic configuration and is highly nonreactive. These belong to Group 18 of the periodic table.

For the given options:

Option A: Hydrogen (H)

The electronic configuration of hydrogen = $1s^1$

This element belongs to Group 1 and is known as alkali metal.

Option B: Barium (Ba)

The electronic configuration of barium = $[Xe]6s^2$

This element belongs to Group 2 and is known as alkaline earth metal.

Option C: Nitrogen (N)

The electronic configuration of nitrogen = $1s^12s^22p^3$

This element belongs to Group 15 and is known as p-block element.

Option D: Krypton (Kr)

The electronic configuration of krypton = $[Ar]4s^23d^{10}4p^6$

This element belongs to Group 18 and is known as noble gas.

Option E: Bromine (Br)

The electronic configuration of bromine = $[Ar]4s^23d^{10}4p^5$

This element belongs to Group 17 and is known as halogen.

Hence, from the above information, the correct answer is Option D.

Novosadov [1.4K]2 years ago

5 0

Krypton is a noble gas! Hope this helps!

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Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficien

Vadim26 [7]

Answer:

_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)

Explanation:

Step 1:

The unbalanced equation:

AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

Step 2:

Balancing the equation.

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)

The above equation can be balanced as follow:

There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:

AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)

There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)

There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:

5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)

Now the equation is balanced

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3 years ago

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Answer:

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Explanation:

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2 years ago

Assume that the test tube shown started out having 10.00 g of mercury(II) oxide. After heating the test tube briefly, you find 1

anyanavicka [17]

This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.

Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

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Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

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