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abruzzese [7]
2 years ago
10

3.4 For each of the following elements, locate it in the Periodic Table, and indicate whether it is a metal, metalloid, or non-m

etal. (3) 4.4.1 Germanium (Ge) 4.4.2 Selenium (Se) 4.4.3 Beryllium (Be)
​
Chemistry
1 answer:
12345 [234]2 years ago
6 0

According to its location in the periodic table:

  • Ge: Period 4, Group 14. Metalloid.
  • Se: Period 4, Group 6, Non-metal.
  • Be: Period 2, Group 2, Metal.

<h3>What is a periodic table?</h3>

It is a table in which elements are ordered in Groups (columns) and Periods (rows).

According to their ubication and properties they can be classified in:

  • Metals: occupy the left and center of the table.
  • Metalloids: occupy the right of the table (dark green in the attached table).
  • Non-metals: occupy the right of the table (yellow and light green in the attached image).

Let's locate and classify the following elements:

  • Germanium: Period 4, Group 14. Metalloid.
  • Selenium: Period 4, Group 6, Non-metal.
  • Beryllium: Period 2, Group 2, Metal.

Learn more about the periodic table here: brainly.com/question/14514242

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An unknown substance has the composition of 77.87% c, 11.76% h and 10.37% o. the compound has a molar mass of 154.25 g/mole. wha
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We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound 
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mass                    77.87 g              11.76 g                 <span>10.37 g
number of moles  77.87/12            11.76/1                 10.37/16
moles                  = 6.48                  = 11.76                 =0.648
divide by least number of moles 
                              6.48/0.648       11.76/0.648            0.648/0.648
                              = 10                 =18.1                      = 1
rounded off
C - 10 , H - 18 and O - 1
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When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver
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Answer:

103.62 g of AgCl.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2

Step 2:

Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.

This is illustrated below:

Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol

Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g

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Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g

Thus, from the balanced equation above,

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Step 3:

Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.

The mass of AgCl produced from the reaction can be obtained as follow:

Form the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.

Therefore, 103.62 g of AgCl were produced from the reaction.

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