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tiny-mole [99]
3 years ago
9

A correct name for fe(no3)2 is __________.

Chemistry
1 answer:
Pani-rosa [81]3 years ago
8 0
Well i believe it is Iron (II) nitrate
You might be interested in
Which reaction occurs spontaneously
coldgirl [10]

Answer:

Chemical Reaction

Explanation:

5 0
3 years ago
Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-
hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written betweenZn^{2+} (Positive ion)and F^{-} (Negative ion)

First Formula would be ZnF_{2}

Second empirical formula is between Zn^{2+}(Positive ion) and O^{2-}(Negative ion)

Second Formula would be Zn_{2}O_{2}

Note : When the subscript are same they get cancel out, so Zn_{2}O_{2} would be written as ZnO

Third empirical formula is between Ni^{4+}(Positive ion) and F^{-}(Negative ion)

Third Formula would be :NiF_{4}

Forth empirical formula is between Ni^{4+}(Positive ion)and O^{2-}(negative ion)

Forth Formula would be : Ni_{2}O_{4} or NiO_{2}

Note- The subscript will be simplified and the formula will be written as NiO_{2}.

The empirical formula of four binary ionic compounds are : ZnF_{2}, ZnO, NiF_{4},NiO_{2}


8 0
3 years ago
Read 2 more answers
A sample of 0.281 gg of an unknown monoprotic acid was dissolved in 25.0 mLmL of water and titrated with 0.0950 M NaOH NaOH. The
tensa zangetsu [6.8K]

Answer:

98.6 g/mol.

Explanation:

Equation of the reaction

HX + NaOH--> NaX + H2O

Number of moles = molar concentration × volume

= 0.095 × 0.03

= 0.00285 moles

By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.

Molar mass = mass ÷ number of moles

= 0.281 ÷ 0.00285

= 98.6 g/mol.

5 0
3 years ago
34g aluminum are combined with 39g chlorine gas which is the limiting reactant
vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

        2 Al + 3 Cl2 ==> 2 AlCl3  represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

                                                = 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

                                                  = 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0
3 years ago
The ksp of yttrium fluoride, yf3, is 8.62 × 10-21. calculate the molar solubility of this compound.
motikmotik

Answer:

The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.

Explanation:

In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.

       YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)

I                       0               0

C                     +S            +3S

E                       S              3S

The solubility product (Ksp) is:

Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴

S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21}  /27}=4.23 \times 10^{-6}M

6 0
4 years ago
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