An ionized has consisting of positive ions and free elections in proportion resulting in more or less no overall electric charge
Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:
Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
Answer: <span>Molecular geometry around each carbon atom in a saturated hydrocarbon is
Tetrahedral.
Explanation: </span> In saturated hydrocarbons (-CH₂-) the central atom (
carbon) is bonded to either three or two hydrogen atoms and one or two carbon atoms. So, the central atom is having four electron pairs and all pairs are bonding pairs and lacks any lone pair of electron. According to
Valence Shell Electron Pair Repulsion (VSEPR)
Theory the central atom with four bonding pair electrons and zero lone pair electrons will attain a
tetrahedral geometry with
bond angles of 109°.
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.
To learn more about precipitate, refer to the below link:
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Answer:
nucleus
Explanation:
The majority of an atoms' mass comes from the protons and neutrons that make up its nucleus. Electrons are the least massive of an atom's constituent particles, with a mass of 9.11 x 10-31 kg and a size too small to be measured by current techniques.