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Bumek [7]
3 years ago
8

What is the valency of Aluminium and nitrate in Al(NO3)3 ?​

Chemistry
1 answer:
Kitty [74]3 years ago
3 0

Answer:xl/ck xLKsx

Explanation:sl/cs c/ls

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Select the correct answer.
Nuetrik [128]

Answer:

A binary covalent compound is composed of two different elements (usually nonmetals). For example, a molecule of chlorine trifluoride, ClF3 contains 1 atom of chlorine and 3 atoms of fluorine.

Rule 1. The element with the lower group number is written first in the name; the element with the higher group number is written second in the name. Exception: when the compound contains oxygen and a halogen, the name of the halogen is the first word in the name.

Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name.

Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the root of the element name (e.g., fluorine = F, "fluoride" = F-; sulfur = S, "sulfide" = S2-).

Rule 4. Greek prefixes are used to indicate the number of atoms of each element in the chemical formula for the compound. Exception: if the compound contains one atom of the element that is written first in the name, the prefix "mono-" is not used.

Explanation:

5 0
3 years ago
Read 2 more answers
Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....
Oduvanchick [21]

Answer:

Explanation:

\mathbf{From \  the  \ information \  given:} \\ \\ \mathbf{The \  rates \  of  \ the \ f ollowing \  reactions \  can \  be \  expressed  \ as \  follows:}

(a)

\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}

(b)

\mathbf{C_2H_6 \to C_2H_4 + H_2}  \\ \\ \\ \mathbf{  -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}

(c)

\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} =  \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt}   }

(d)

\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\  \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}

(e)

\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\  \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}

5 0
3 years ago
Which of the following molecules is nonpolar?
Dvinal [7]

Answer:

CO2

Explanation:

  • There are two types of molecules
  1. Polar
  2. Non polar

Non polar molecules are insoluble in water .

7 0
3 years ago
In the redox conversion of Ni2+ to NiO4−, the oxidation number of Ni goes from (−2, 0, +2) to (−1, +1, +7, +9). Recall that the
inessss [21]
Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero, 
for NiO4-, assuming oxidation number of Ni is x 
 (x ×1) + (-2 × 4) = -1
  x = + 7
Therefore, the oxidation number goes from 0 to +7
7 0
3 years ago
What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl
fenix001 [56]
M(P)=3.72 g
M(P)=31 g/mol

m(Cl)=21.28 g
M(Cl)=35.5 g/mol

n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol

n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol

P : Cl = 0.12 : 0.60 = 1 : 5

PCl₅ - is the empirical formula
6 0
3 years ago
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