When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.
<h3>What is a precipitate?</h3>
The precipitate is the solid concentration of a substance that is collected over a solution.
First, we determine the concentration of magnesium and hydroxide
(Mg2+) = 7.00 × 10⁻⁴
(OH−) = 5.00 × 10⁻⁵
Now, we calculate the solubility quotient
Qc = (Mg2+) (OH−) ²
Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²
Qc = 1.75 x 10⁻¹²
The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.
Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.
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Answer:
1360kJ are evolved
Explanation:
When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.
To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:
<em>Moles H2 -Molar mass: 2g/mol-</em>
90g H2 * (1mol / 2g) = 45 moles
<em>Moles O2 -Molar mass: 32g/mol-</em>
90g * (1mol / 32g) = 2.81moles
For a complete reaction of 2.81 moles of O2 are needed:
2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2
As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.
As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:
2.81 moles O2 * (241.8kJ / 1/2moles O2) =
<h3>1360kJ are evolved</h3>
Complete Question:
Suppose a cobalt atom in the +3 oxidation state formed a complex with two bromide (Br-) anions and four ammonia (NH3) molecules. write the chemical formula of this complex.
Answer:
[Co(NH₃)₄]⁺Br₂
Explanation:
The cobalt atom with +3 oxidation is represented as Co⁺³, and if it's bonded to two bromide ions, and four ammonia molecules. The molecules that are bonded to the metal atom (Co) are called complexing agents.
In the representation, we first put the molecules that surround the metal atom, forming an anion with the oxidation of the metal:
[Co(NH₃)₄]⁺³
Then, the ions are put in the formula. Because there are two bromides ion, each one with 1 minus charge, only 2 plus charged will be neutralized, and the complex will be:
[Co(NH₃)₄]⁺Br₂
