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Paul [167]
3 years ago
5

Three useful applications of alkaline earth metals

Chemistry
1 answer:
svetlana [45]3 years ago
7 0
Alkaline earth metals refer to a group of elements in the periodic table. They include beryllium, magnesium, calcium, strontium, barium, and radium. Main uses of these elements are given below. Beryllium: It is used in small amounts to make copper beryllium alloy that are very strong and hard.
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It’s either a or b but I’m not sure
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How many moles of Ar atoms are there in 202 L of Ar gas at STP? A) 1.20 x 1024 mol B) 1.11 x 10-1 mol C) 4.52 x 103 mol D) 9.02
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Answer:

4.52 x 103 mol

Explanation:

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Show with dot and cross diagram of the formation of ions in the reaction of Fluorine with sodium ions.
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Answer:

Electron dot diagram is attached below

Explanation:

Sodium is alkali metal and present in group one. It has one valence electron. All alkali metal form salt when react with halogens.

Sodium loses its one electron to get stable.  While all halogens have seven valence electrons they need only one electron to get stable electronic configuration.

When alkali metals such as sodium react with halogen fluorine it loses its one valence electron which is accepted by fluorine and ionic bond is formed. The compound formed is called sodium fluoride.

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Electron dot diagram is attached below.

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4 years ago
The structure of cyclopentane-1,2-diol​
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3 years ago
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

  • 1.67 mol * 65 g/mol = 108.6 g
6 0
3 years ago
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