Answer:
0.0428 M
Explanation:
Because we're asked to calculate the molarity of nickel(II) cation, we need to <u>determine all sources for that species</u>, in this case, all Ni⁺² comes from the nickel(II) bromide solid (NiBr₂).
We use the molecular weight of NiBr₂ to calculate the moles of Ni:
1.87 g NiBr₂ ÷ 218.49g/mol * (1molNi⁺²/1molNiBr₂) = 8.55x10⁻³ mol Ni⁺²
Then we <u>divide the moles by the volume in order to calculate the concentration</u>:
8.55x10⁻³ mol Ni⁺² / 0.200 L = 0.0428 M
Explanation:
The enzyme 's active site binds to the substrate. Increasing the temperature generally increases the rate of a reaction, but dramatic changes in temperature and pH can denature an enzyme, thereby abolishing its action as a catalyst. ... When an enzyme binds its substrate it forms an enzyme-substrate complex.
- friend,please mark my answer in brainliest answers
- friend,please follow me
- friend,please thanks this answer
- friend,please vote it 5 star
Answer:
B. oxidation; reduction
Explanation:
A voltaic cell is electro-chemical cell in which chemical energy is converted into electrical energy.
1. This cell utilizes chemical reaction to generate electric.\
2. there two electrode anode and cathode
3. At Anode oxidation occurs
4. At cathode reduction occurs
5. chemical is present in the cell which is electrolyte which completes the circuit of the voltaic cell.
- oxidation is the process in which there is loss of electrons
- Reduction is the process in which there is gain of electrons
___________________________________________________
Based on above discussion
At anode oxidation takes place
At cathode reduction takes place.
Hence, correct option is B. oxidation; reduction
Answer:
it answer D because waste is flowing thru the air
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
