An atom of an element with two valence electrons reacts with two atoms of an element with atomic number 17 to form an ionic compound. What is the compound MOST LIKELY formed as described? A) CaF2 B) MgCl2 C) CO2 D) BeBr2


Answer 1

Answer is: B) MgCl2.

Chlorine has atomic number 17, it means it has 17 protons and 17 electrons.

Electron configuration of chlorine atom: ₁₇Cl 1s² 2s² 2p⁶ 3s² 3p⁵.

Chloride is negative ion of chlorine. Chloride is formed when chlorine gain one lectron.

Chloride anion has 17 protons and 18 electrons (like argon-noble gas).

The electron configuration for the chloride ion: ₁₇Cl⁻ 1s²2s²2p⁶3s²3p⁶.

Magnesium has a 2+ oxidation number and chlorine has a 1- oxidation number.  

Magnesium is metal from second group of Periodic table of elements and it lost two electrons to have electron configuration as closest noble gas neon (₁₀Ne), chlorine is nonmetal from 17. group of Periodic table and it gains one electron to have electron configuration as argon (₁₈Ar).

Answer 2
Answer: The answer will be B) MgCl2 because magnesium has 2 valence electrons and is a metal and chlorine has the atomic number of 17 and is a metal and in order to form an ionic compound, you must have 1 metal and 1 nonmetal

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It's believed by many that a good way to vandalize a car is to pour sugar (sucrose) in its gas tank. the sugar is believed to dissolve in the gasoline, enter the gas lines and then "caramelize" into a gunky texture when combusted. if you wanted to test this theory casually, without jeopardizing anyone's car, which of the solvents used in the experiment would be the best to use?


Answer is: in this experiment it is best to use some liquid alkanes (for example hexane), because gasoline consists of hydrocarbons with between four and twenty carbon atoms. Gasoline is a mixture  of many different hydrocarbons: alkanes (paraffins), cycloalkanes  and alkenes (olefins).

Calculate the molarity of a solution that contains 0.175 mol of zncl2 n exactly 150 ml of solution.



Molarity is the number of moles present in liter of a solution.

Mathematically,    Molarity =

It is given that no. of moles present into the solution are 0.175 mol and volume is 150 ml.

As 1 ml = 0.001 L. So, 150 ml will be equal to 0.15 L.

Hence, calculate the molarity as follows.

    Molarity =


                  = 1.16 M

Thus, we can conclude that molarity of the given solution is 1.16 M.

Molarity = mol/L

0.175 mol
0.15 L (1000 mL = 1 L)

Molarity = 0.175 mol/0.15 L = 1.1666667 M

If body fats has a density of 0.94 g/mL and 3.0 liters of fat are removed, how many pounds of fat were removed from the patient?


The mass of the fat removed from the patient is 6.34 pounds

To calculate the mass of the fat (in pounds) removed  from the patient, we use the formula below


  • D = m/V.............. Equation 1


  • D = Density of the body fat
  • V = volume of the body fat
  • m = mass of the body fat.

make m the subject of the equation.

  • m = DV.................. Equation 2

From the question,


  • D = 0.96 g/mL
  • V = 3 L = 3000 mL

Substitute these values into equation 2

  • m =0.96(3000)
  • m = 2880 g

The convert mass from gram to Ponds

  • m = (2880/454) Pounds = 6.34 pounds.

Hence, The mass of the fat removed from the patient is 6.34 pounds

Learn more about mass here:


6.217 pounds


We are given;

  • Density of body fats 0.94 g/mL
  • Volume of fats removed = 3.0 L

We are required to determine the mass of fats removed in pounds.

We need to know that;

Density = Mass ÷ volume

1 L = 1000 mL, thus, volume is 3000 mL

Rearranging the formula;

Mass = Density × Volume

         = 0.94 g/mL × 3000 mL

         = 2,820 g

but, 1 pound = 453.592 g


Mass = 2,820 g ÷ 453.592 g per pound

         = 6.217 pounds

Thus, the amount of fats removed is 6.217 pounds


How many half-lives will it take for the concentration of the n2o to reach 30 % of its original concentration?



Number of lives required to reach 30% of the its original concentration is 2.


Number of half lives are calculated by using formula :



a = amount of reactant left after n-half lives

= Initial amount of the reactant

n = number of half lives

Here we have:

Initial concentration of nitrogen dioxide =

Concentration of nitrogen dioxide after n number of half lives = a

a = 30% of x :

n = 1.736 ≈ 2

Number of lives required to reach 30% of the its original concentration is 2.

We have that after 1 half-life, the reactant reaches half its concentration . After 2 half-lives it reaches of its concentration, namely one-fourth. After n half lives we have that it reaches of its initial concentration. Thus, we have to solve the equation . Taking the log base 2 of each part of the equation we get from logarithm properties: . The other hand yields: . Solving for n we get that n=1.74. Thus, after around 1.74 half lives the concentration becomes 30% of the initial one. Note how this is consistent with our previous analysis; if we let it a little bit more, 2 half-lives, the concentration will become 25% of the initial.
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