Answer:
- See the image attached. It is taken from an online chemistry textbook.
- See the explanation below.
Explanation:
<em>Sodium chloride</em> consits of sodium cations (positive ions), Na⁺, and chloride anions (negative ions), CL⁻.
<u>Pure sodium chloride</u> is packed in crystals: sodium ions and chloride ions are packed together and the ions are in fixed positions. There are not free electrons that can move. Thus, sodium chloride doesn't conduct electricity, because there are no electrons or ions which are free to move.
In aqueous solution, sodium chloride units dissociates into their ions:
Those ions are freely to move in the solution, and such they are charge carriers, which conduct the electricity.
As explained above, in solid sodium chloride, the ions cannot move and there is not flow of current.
That is why solid pure salt of NaCl does not conduct electricity and the solutions of NaCl do conduct electricity.
The image attached show both diagrams. In the diagram A, the ions are packed together, showing that they cannot move. In the diagram B, the ions are dissolved in water, showing that they can move and carry the charge, allowing the flow of current.
For this type of problem, it is essential for you to have a data on the standard heats of formation of the substances given. For elemental substances or diatomic gases, the standard heat of formation is 0. Standard means the temperature is at 0°C and pressure at 1 atm. Calculate the standard heat of reaction using:
ΔH°rxn = ∑(Stoichiometric coefficient×ΔHf of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)
Then, use this equation to find the reaction at T = 500°C and P = 1 bar:
ΔHrxn = ΔH°rxn + [∑(Stoichiometric coefficient×Cp of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)]ΔT
So, you also need the Cp or specific heat capacities of the substances.
Answer:
A. The farther an electron is from the nucleus, the greater its energy.
Explanation:
To help, I drew a diagram. This represents an ionic bond between Na and Cl. Na is giving his single electron to Cl, which is indicated by the arrow, to make Cl full with 8 electrons.
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
