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lakkis [162]
3 years ago
15

You want to analyze a cadmium nitrate solution. what mass of naoh is needed to precipitate the cd2+ ions from 32.7 ml of 0.499 m

cd(no3)2 solution?
Chemistry
1 answer:
kodGreya [7K]3 years ago
6 0
First, we must know that in order to remove one mole of cadmium ions, we will require two moles of sodium ions. This is visible from the fact that the valence charge of cadmium ions is +2, while for sodium ions it is +1.

Next, we use the equation:

Moles = Molarity * Liters

We know that:

moles of sodium =  2 * moles of cadmium

Now, the moles of cadmium present are:
Moles = 0.499 * 0.0327 = 0.0163

The moles of NaOH required are double this, so:

Moles NaOH = 2 * 0.0163 = 0.0326

Mass = moles * molecular weight

Mass  = 0.0326 * 40
Mass = 1.304 grams of NaOH
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It is important to have the correct bond angles of the different atoms and the shape of the molecule due to following reasons;

                  Among other properties the polarity of compounds mainly depend upon the shape and bond angles of that particular compound. For example, considering the molecule of water, we already know that it is a polar molecule with partially positive hydrogen atoms and partially negative oxygen atoms and acts as universal solvent. The bond angle in water is about 104.5° with a Bent geometry. Unlike carbon dioxide (CO₂) which has Linear structure with bond angle 180° and is non-polar in nature therefore, the bent geometry in water is responsible for the polarity.

                  Other properties which can also be predicted by predicting the bond angles along with molecular geometries are;

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What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl
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Answer:

\large \boxed{\text{0.012 mol}}  

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

               CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL:                    70.

c/mol·L⁻¹:             0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }

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