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lakkis [162]
4 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]4 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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When you count up the electrons in a Lewis Structure, do double and triple bonds count as 4 or 6 electrons?
natima [27]

Answer:

Double=4 and triple=6

Explanation:

This is because double bonds are two pairs of electrons are shared between atoms and triple bonds are three pairs, and one pair of electrons is 2, so 2 x 2=4 and 2 x 3=6.

6 0
3 years ago
What is the difference between elimination and substitution reaction
jasenka [17]

Answer:

a) E2

b) SN2

c) SN2

Explanation:

A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.

We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.

7 0
3 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

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