It is in a large elliptical shape!
Answer:
4d orbital.
Explanation:
Hello!
In this case, since zirconium's atomic number is 40, we fill in the electron configuration up to 40 as shown below:
Thus, the orbital 4d is partially filled.
Best regards!
Answer: Reducing agent in the given reaction is 
.
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in is +2 and 
has 0 oxidation state.
In 
oxidation state of S is 2.5 and in 
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is .
Answer:
The answer to your question is 80.3%
Explanation:
Data
Percent by mass of F
molecules NF₃
Process
1.- Calculate the molar mass of nitrogen trifluoride
molar mass = (1 x 14) + (19 x 3)
= 14 + 57
= 71 g
2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.
71 g of NF₃ ------------------ 100%
57 g of F ------------------- x
x = (57 x 100)/71
x = 5700 / 71
x = 80.3%
3.- Conclusion
Fluorine is 80.3% by mass of the molecule NF₃
Answer:
M HCl sln = 12.0785 M
Explanation:
- molarity (M) [=] mol/L
- %mm = ((mass compound)/(mass sln))*100
∴ mass sln = 100.0 g
∴ δ sln = 1.19 g/mL
∴ % m/m = 37 %
⇒ 37 % =((mass HCl/mass sln))*100
⇒ 0.37 = mass HCl / 100.0 g
⇒ 37 g = mass HCl
∴ molar mass HCl = 36.46 g/mol
⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol
⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L
⇒ M HClsln = 1.015 mol/0.084034 L
⇒ M HCl sln = 12.0785 M
