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Blababa [14]
3 years ago
14

Would a precipitate form if a solution contained 8.0 x 10-4 m silver nitrate and 1.8 x 10-4 m potassium chromate? you will have

to find ksp values from your text or from another source.
Chemistry
1 answer:
Alina [70]3 years ago
6 0
Chemical reaction: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄<span>(s).
</span>Ksp(Ag₂CrO₄) = 1,1·10⁻<span>¹².
</span>[Ag⁺] = c(AgNO₃) = 8.0·10⁻⁴ M.
[CrO₄²⁻] = c(K₂CrO₄) = 1.8·10⁻⁴ M.
Q(Ag₂CrO₄) = [Ag⁺]²·[CrO₄²⁻]. 
Q = (8.0·10⁻⁴ M)² · 1.8·10⁻⁴ M.
Q = 1.152·10⁻¹⁰.
Q<span> > Ksp, </span><span>a precipitate of silver chromate will form.
</span>Ksp is the solubility product constant for a solid substance dissolving in an aqueous solution.

<span> [Ag</span>⁺<span>] is equilibrium concentration of silver cations.
[CrO</span>₄²⁻] is equilibrium concentration of chromate anions.


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6 0
2 years ago
The bonds in the reactants of Figure 7-3 contained 372 kJ of chemical energy and the bonds in the products contained 350 kJ of c
satela [25.4K]

Answer: 22 kJ amount of energy is released in the following reaction.

Explanation: There are two types of reaction on the basis of amount of heat absorbed or released.

1. Endothermic reactions: These are the type of reactions in which reactants absorb heat to form the products. The energy of the reactants is less than the energy of the products.

2. Exothermic reactions: These are the type of reactions in which heat is released from the chemical reactions. The energy of the products is less than the reactants.

Sign convention for \Delta H: This value is negative for exothermic reactions and positive for endothermic reactions.

For the given chemical reaction,

Energy of the products is less than the energy of the reactants, Hence, this reaction will be a type of exothermic reaction and energy will be released during this chemical change.

Amount of energy released = (350 - 372) kJ = -22kJ

Negative sign symbolizes the energy is being released. So, 22 kJ amount of energy is released in the following reaction.

6 0
2 years ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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