<u>Answer:</u> The complete molecular, ionic, and net ionic equations are given below. The spectator ions are sodium and nitrate ions.
<u>Explanation:</u>
The ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.
The net ionic equation is defined as the equations in which spectator ions are not included.
Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.
The balanced molecular equation for the reaction of lead (II) nitrate and sodium sulfide follows:
The ionic equation follows:
As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.
The net ionic equation follows:
It helps to map out how you will navigate through your unit analysis problem before setting it up.
You are given moles and need grams. What can be used as a conversion factor from moles to grams? Molar mass. We are working with aluminum, so we will need the molar mass of aluminum. My Periodic Table tells me the molar mass of aluminum is approximately 27 g/mol. Now we are ready to set up the unit analysis.
Moles must go on the bottom so that they cancel. Notice how our number of significant figures is 2, so the answer must round to 16 g Al.
<h3>
Answer:</h3>
16 grams
Answer:
E. potassium (K) and bromine (Br)
Explanation:
An ionic bond is formed between compounds with a large electronegativity difference between them. It is usually between a metal and non-metal.
- Potassium is a true metal found in group 1 on the periodic table.
- Bromine is a highly electronegative non-metal which is a halogen.
- Potassium will lose one of its electrons which will be gained by the Bromine.
- The electrostatic attraction between the two species will cause the ionic bond to form.
- The ability of one specie willing to lose electron and the other gaining, is the main bed rock of ionic bonding.
Answer:
A. The reaction will proceed forward forming more CH4
B. The reaction will proceed forward forming more CH4
C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.
D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4
E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.
F. The following will favour CH4 at equilibrium
i. Catalyst to the reaction mixture,
ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture
iii. Adding more C to the reaction mixture.
