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kiruha [24]
3 years ago
14

What observations support the assumption that all of the iodine reacted with the zinc explain?

Chemistry
1 answer:
natita [175]3 years ago
5 0

Answer: It will lead to form a white colored product ZnI_2.

Explanation: The reaction of iodine and zinc involves the direct reaction between a metal and a non-metal leading to the synthesis of a metal salt.

Zn+I_2\overset{H_2O}{\rightarrow}ZnI_2(aq)\overset{\Delta }{\rightarrow}ZnI_2(s)

During this reaction, heat is released (exothermic reaction) while forming Zinc Iodide(ZnI_2) which can be easily obtained by evaporating the solvent.

By looking at the stoichiometric coefficients of reactants and products, we observe that whole of the iodine has reacted with zinc.

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How are quantum numbers related to atomic orbitals?
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According to the Arrhenius concept, which of the following substances is NOT a base in aqueous solutionsA. NH3B. LiOHC. NaOHD. H
olga_2 [115]

Answer:

NH₃ (Option A)

Explanation:

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It can be used only with compounds with H, or OH.

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4 years ago
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3 years ago
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A compound is found to contain 37.32 % phosphorus , 16.88 % nitrogen , and 45.79 % fluorine by
Alla [95]

Answer: 1. The empirical formula is PNF_2  

2. The molecular formula is PNF_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P =\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles

Moles of N =\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles

Moles of F =\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = \frac{1.20}{1.20}=1

For N = \frac{1.20}{1.20}=1

For F =\frac{2.41}{1.20}=2

The ratio of P: N: F= 1: 1: 2  

Hence the empirical formula is PNF_2

The empirical weight of PNF_2= 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1

The molecular formula will be=1\times PNF_2=PNF_2

3 0
3 years ago
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