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Gre4nikov [31]
2 years ago
7

Tetraphosphorus decaiodide is represented by which formula? A) P3I10 B) P4I10 C) P3I8 D) P4I8

Chemistry
2 answers:
dusya [7]2 years ago
6 0

Answer: B) P_4I_{10}

Explanation: P_4I_{10} is a covalent compound because sharing of electrons takes place between phosphorous and iodine. Both the elements are non-metals and hence, will form covalent bond.

The nomenclature of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written then, and a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms , 'tetra' for 4 atoms and so on.

So, the IUPAC name for P_4I_{10} is tetraphosphorus decaiodide.

oksano4ka [1.4K]2 years ago
3 0
The answer is B.

Tetraphosphorus is written as P4, and decaiodide as I10, so when you put them together, it's P4I10
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olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = H_{picric}

So,  H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

The ICE table can be given as:

                          H_{picric}     +       H_2}O          ⇄      H_3}O^+     +     Picric^-

Initial:                0.52                                               0                  0

Change:             - x                                                 + x                 + x

Equilibrium:      0.52 - x                                        + x                 + x

Given that;

acid dissociation constant  (K_a) = 0.42

K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}

0.42 = \frac{[x][x]}{0.52-x}}

0.42 = \frac{[x]^2}{0.52-x}}

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x²  + 0.42x - 0.2184 = 0                   -------------------- (quadratic equation)

Using the quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}    ;     ( where +/-  represent ± )

= \frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}

= \frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}

= \frac{-0.42+\sqrt {1.0496} }{2}     OR   \frac{-0.42-\sqrt {1.0496} }{2}

= \frac{-0.42+1.0245}{2}       OR    \frac{-0.42-1.0245}{2}

= \frac{0.6045}{2}                 OR    -\frac{1.4445}{2}

= 0.30225          OR     - 0.72225

So, we go by the +ve integer that says:

x =  0.30225

x = [ H_3}O^+ ] = [   Picric^- ] =  0.3023  M

∴  the value of  [H3O+] for an 0.52 M solution of picric acid  = 0.3023 M     (to 4 decimal places).

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2 years ago
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2 years ago
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Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
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Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
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Answer:

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Explanation:

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