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3 years ago

Which element in the third period would you expect to have the larger atomic radius, sodium (Na) or Sulfur (S)?

Chemistry

2 answers:

jeyben [28]3 years ago

7 0

Answer: The correct answer is Option B.

Explanation:

The given elements that is sodium and sulfur are the elements which belong to the same period that is third period.

As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.

Therefore, the correct answer is Option B.

Doss [256]3 years ago

3 0

Answer:

Explanation:

As, the electrons are added up to the same shell across a period, the number of protons also increase and hence, they attract the electrons more effectively. This means that the effective nuclear charge also increases and the size of the sulfur thus shrink and hence, the size (atomic radius) of sodium is the largest amongst third period.

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Prepare a 1.00 L of M solution of sodium carbonate . Determine the amount of sodium carbonate needed incorrect significant figur

Vedmedyk [2.9K]

Answer:b

Explanation:

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3 years ago

Write the symbolic notation of an isotope of an element having 8 protons, 8 electrons, and 11 neutrons. Click on the “Templates”

horrorfan [7]

Attached to this answer is the format of Isotope Notation that you can use for future reference. (Please open)

There are 8 Protons. The Atomic Number is the same number of an element's proton.

If you can see in the format, the mass number is calculated by adding the atomic number/protons and neutrons.

Mass number = 8 + 11
Mass number = 19

The image of the final answer is attached as well.

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3 years ago

What is a type of air mass that forms over oceans?

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The answer is maritime air masses

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3 years ago

Read 2 more answers

A student has a balloon with a volume of 2.5 liters that contains 4.0 moles of air. The ballon has a small leak, allowing one mo

Over [174]

Answer:

1.9 L

Explanation:

Step 1: Given data

Initial number of moles of air (n₁): 4.0 mol

Initial volume of the balloon (V₁): 2.5 L

Final number of moles of air (n₂): 3.0 mol

Final volume of the balloon (V₂): ?

Step 2: Calculate the final volume of the balloon

According to Avogadro's law, the volume of an ideal gas is directly proportional to the number of moles. We can calculate the final volume of the balloon using the following expression.

V₁ / n₁ = V₂ / n₂

V₂ = V₁ × n₂ / n₁

V₂ = 2.5 L × 3.0 mol / 4.0 mol

V₂ = 1.9 L

6 0

2 years ago

How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has

Papessa [141]

The balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas= $\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.

number of moles of sodium azide = $16 moles$

Mass of sodium azide in g = $number of moles\times molar mass of sodium azide$ .

= $16 moles\times 65.00 g/mol$

= $1040 g$

Thus, mass of sodium azide which is required to produce $19.0 ft^{3}$ of nitrogen gas = $1040 g$

3 0

2 years ago