Answer:
C3 H6 O2
Explanation:
first divide their mass by their respective molar mass, we get:
30.4 moles of C
61.2 moles of H
20.25 moles of O
now divide everyone by the smallest one of them then we get
C= 1.5
H= 3
O= 1
since our answer of C is not near to any whole number so we will multiply all of them by 2
so,
C3 H6 O2 is our answer
Answer: Molarity of
anions in the chemist's solution is 0.0104 M
Explanation:
Molarity : It is defined as the number of moles of solute present per liter of the solution.
Formula used :

where,
n= moles of solute
= volume of solution in ml = 100 ml
Now put all the given values in the formula of molarity, we get

Therefore, the molarity of solution will be 

As 1 mole of
gives 2 moles of 
Thus
moles of
gives =
Thus the molarity of
anions in the chemist's solution is 0.0104 M
Answer: Alex should get 300 Calories from carbohydrates, which is about 33 grams. This snack is almost 70% of her total grams of carbohydrates per day.
Answer:
a) Mo the electron configuration: 42Mo: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d4
Mo3+ - is Paramagnetic
b) Au - [Xe] 4f14 5d10 6s1
For Au+ is not paramagnetic
c) Mn - [Ar] 3d5 4s2
Mn2+ is paramagnetic
d)Hf -[Xe] 4f¹⁴ 5d² 6s²
Hf2+ is not paramagnetic
Explanation:
An atom becomes positively charged when it looses an electron.
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
Answer:

Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
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