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viktelen [127]
3 years ago
7

Which nonmetal is the "wrong" place on the periodic table?

Chemistry
2 answers:
Tanya [424]3 years ago
8 0
There is no picture so I can't say the answer please update your post with the image BECUASE all non metals are located in the same group in the periodic table
bixtya [17]3 years ago
4 0

Answer:

The correct answer is: Hydrogen

Explanation:

The periodic table is defined as the tabular arrangement and classification of the chemical elements. This table contains seven rows, called periods; and eighteen columns, called groups.

Generally, the <u>metals are present on the left</u> and the <u>non-metals are present on the right</u> of the periodic table.

However, Hydrogen (atomic number 1) is the only <u>non-metal </u>that is placed in <u>group 1, on the left-side of the periodic table</u>. This is because according to the electronic configuration of hydrogen (1s¹), it should be placed in group 1 of the periodic table.

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Which of these solids is insoluble in water?
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Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?
Andrews [41]
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 Zn+2(?M) + 2e- === Zn(s) (cathode)
 
 Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn 
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 0.023 = -0.0296 { log 0.10 – log [Zn+2] }
 0.023 = -0.0296{ -1 - log[Zn+2] }
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6 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0
3 years ago
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