Answer:
: the answer is C
Explanation:
:State the law of conservation of energy.
Define and endothermic process.
Define an exothermic process.
Make conversions involving heat units.
Answer:
sp³d¹ hybridization
Explanation:
Given Cl as central element with three F substrates ...
The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.
Valence bond theory predicts the following during bonding:
Cl:[Ne]3s²3p²p²p¹3d⁰
=> [Ne]3s²p²p¹p¹d¹
=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹
giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.
Note the following images:
Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.
Answer:
Water has strong hydrogen bond dipole-dipole intermolecular forces that give water a high surface tension and a high heat of vaporization and that make it a strong solvent.
Explanation:
Answer:
5.25 moles.
Explanation:
The decomposition reaction of NaN₃ is as follows :

We need to find how many grams of N₂ produced in the process.
From the above balanced chemical reaction, we conclude that the ratio of moles of sodium azide and nitrogen gas are 2 : 3.
2 moles of sodium azide decomposes to give 3 moles of nitrogen gas. So,
3.5 moles of sodium azide decomposes to give
moles of nitrogen gas.
Hence, the number of moles produced is 5.25 moles.
Answer:
Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)
Explanation:
Part 1. Volume of reactant
(a) Balanced chemical equation.

(b) Moles of CuCl₂

(c) Moles of Na₃PO₄
The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

(d) Volume of Na₃PO₄

Part 2. Net ionic equation
(a) Molecular equation

(b) Ionic equation
You write molecular formulas for the solids, and you write the soluble ionic substances as ions.
According to the solubility rules, metal phosphates are insoluble.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>
The net ionic equation is
3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)