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A 25.0 ml sample of an unknown monoprotic acid was titrated with 0.12 m naoh. the student added 31.6 ml of naoh and went past th

Inessa05 [86]

The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants.
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.

7 0

3 years ago

When 240 mg of a certain molecular compound X are dissolved in 35.0 g of dibenzyl ether ((C6H5CH2)2O), the freezing point of the

DedPeter [7]

Answer: MM = 16.55 g/mol

Explanation: Freezing point depression is a phenomena that explains why adding a solute to a solvent decreases the solvent freezing point: when a substance begins to freeze, its molecules slows down and rearrange itself forming a solid. If a solute is added, the molecules from the solvent interfere in the formation of the solid. To guarantee the transformation, the solution has to cooled down even more.

Freezing point and molality concentration is related by

$\Delta T=T_{f}_{(solvent)}-T_{f}_{(solution)}=K_{f}.m$

where

ΔT is freezing point depression

$T_{f}_{(solvent)}$ and $T_{f}_{(solution)}$ are freezing point of solvent and solution, respectively

$K_{f}$ is freezing point depression constant

m is molality concentration

Dibenzyl ether is the solvent and has the following properties: $K_{f}=$ 6.27 and $T_{f}$ = 3.6°C.

Molality concentration is

$m=\frac{T_{(solvent)}-T_{(solution)}}{K_{f}}$

$m=\frac{3.6-1}{6.27}$

m = 0.415

Molality concentration is moles (n) of solute dissolved in a mass, in kilogram, of solvent.

$m=\frac{moles}{mass(kg)}$

n = m(mass of solvent in kg)

n = 0.415(0.035)

n = 0.0145

Molar mass (M) is the weight of one sample mole and can be calculated as

$n=\frac{m}{M}$

M = $\frac{m}{n}$

m in grams

Molar mass of compound X is

$M=\frac{0.24}{0.0145}$

M = 16.55

Molar mass of molecular compound X is 16.55g/mol

3 0

3 years ago

Find the percentage composition of each element in the compound having 9.8 g of nitrogen 0.7 g of hydrogen and 33.6 g of oxygen

Rainbow [258]

Answer:

Search Results

Empirical formula of carbon hydrogen and nitrogen

Nitrogen usually forms three bonds one to Carbon and 2 to Hydrogen atoms. Carbon usually forms four bonds one to Nitrogen and 3 to Hydrogen atoms. The empirical formula is CH5N .

Explanation:

4 0

4 years ago

How much heat is absorbed during production of 147 g of NO by the combination of nitrogen and oxygen?

marin [14]

Answer:

$\large \boxed{\text{105 kcal}}$

Explanation:

MM: 30.01

N₂ + O₂ ⟶ 2NO; ΔH = +43 kcal/mol

m/g: 147

Treat the heat as if it were a reactant in the reaction. Then you can write

N₂ + O₂ + 43 kcal ⟶ 2NO

The conversion factor is then 43 kcal/2 mol NO.

1. Moles of NO

$\text{Moles of NO} = \text{147 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{4.898 mol NO}$

2. Amount of heat

$\text{Heat} = \text{4.898 mol NO } \times \dfrac{\text{43 kcal}}{\text{2 mol NO}} = \text{105 kcal}\\\\\text{The reaction absorbs $\large \boxed{\textbf{105 kcal}}$}$

7 0

3 years ago

What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigo- nal planar arra

Reika [66]

Answer:

sp³;

sp²;

sp;

None;

One;

Two;

They're used to pi bonds.

Explanation:

The central atom in a molecule is generally the one that can make a greater number of bonds. The covalent bonds are made by the sharing of electrons, and, for that, the electron must be alone in the orbital.

To explain this, the hybridization theory was created, which states that, the orbitals are joined to form hybrids ones, and so, by the Hund's law, the electrons are alone in them.

The sigma bonds are done in the hybrids orbitals, and at the pure orbitals, the pi bonds are done. The lone pair of electrons are at a pure orbital. So, to know the hybridization of the central atom, we must know how many sigma bonds it does, and it will be the number of hybrids orbitals (each orbital may have two electrons, thus each bond are done in one orbital).

Double bonds and triple bonds have always only one sigma bond, so the number of sigma bonds is equal to the number of bonds, it's not necessary to know if they are simple, double or triple.

When the arrangement is tetrahedral, the central atom does 4 bonds, so it has 4 sigma bonds, and 4 hybrids orbitals (one of s and three for p), does its hybridization is sp³. Because exists only 3 p orbitals, there are no unhybridized p orbitals in this case.

When the arrangement is trigonal, the central atom does 3 bonds, so it has 3 hybrids orbitals (one of s and two of p), thus the hybridization is sp². So there are one unhybridized p orbitals.

When the arrangement is linear, the central atom does 2 bonds, so it has 2 hybrids orbitals (one of s and one of p), thus the hybridization is sp. So, there are two unhybridized p atoms.

As stated before, the unhybridized p orbitals are used to pi bonds.

5 0

3 years ago