Answer:
They have electrons in their 3d- and 4s-orbital for bond formation.
Explanation:
d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.
The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.
If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.
If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2
Answer: a. 0.26mol
b. 0.000479mol
c. 1.12mol
Explanation: Please see attachment for explanation
<u>Answer:</u> The percentage abundance of 
and 
isotopes are 75.77% and 24.23% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of isotope be 'x'. So, fractional abundance of isotope will be '1 - x'
- <u>For isotope:</u>
Mass of isotope = 34.9689 amu
Fractional abundance of isotope = x
- <u>For isotope:</u>
Mass of isotope = 36.9659 amu
Fractional abundance of isotope = 1 - x
- Average atomic mass of chlorine = 35.4527 amu
Putting values in equation 1, we get:
![35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577](https://tex.z-dn.net/?f=35.4527%3D%5B%2834.9689%5Ctimes%20x%29%2B%2836.9659%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.7577)
Percentage abundance of isotope = 
Percentage abundance of isotope = 
Hence, the percentage abundance of and isotopes are 75.77% and 24.23% respectively.
The overall balanced reaction equation is;
4Zn(s) + 10H^+(aq) + NO3^-(aq) -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l)
<h3>What is the balanced reaction equation?</h3>
The redox reaction equation is said to be balanced when the number of electron gained is equal to the number of electrons lost.
Now;
1. Reduction and oxidation half-reactions
Zn(s) -----> Zn^2+(aq) + 2e
And
NO3^-(aq) ---->NH4^+(aq) + 3H2O(l)
2. Using the H2O and H+ to balance O and H;
4Zn(s) + 10H^+(aq) + NO3^-(aq) -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l)
3. Balancing the electrons lost and gained; 4Zn(s) + 10H^+(aq) + NO3^-(aq) + 8e -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l) + 8e
4. The overall balanced reaction equation is;
4Zn(s) + 10H^+(aq) + NO3^-(aq) -----> 4Zn^2+(aq) + NH4^+(aq) + 3H2O(l)
Learn more about redox reaction:brainly.com/question/13293425
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