<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.
<u>Explanation:</u>
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.
The given redox reaction follows:

To balance the given redox reaction in basic medium, there are few steps to be followed:
- Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons
Oxidation half-reaction: 
Reduction half-reaction: 
- Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions
Oxidation half-reaction:
( × 3)
Reduction half-reaction:
( × 2)
The half-reactions now become:
Oxidation half-reaction: 
Reduction half-reaction: 
- Add the equations and simplify to get a balanced equation
Overall redox reaction: 
As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution
To measure the distance between two opposite sides of an object
Answer:
81.59%
Explanation:
First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:
- 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃
Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:
- 6.32 mol NH₃ *
= 6.32 mol NO
Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:
- 6.32 mol NO * 30 g/mol = 189.60 g NO
Finally we <u>calculate the percent yield</u>:
- 154.70 g / 189.60 g * 100% = 81.59%
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
Answer:
positive force → balanced force → negative force
Explanation:
np