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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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"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"
Artyom0805 [142]

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-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

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4 0
3 years ago
At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

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