The maximum amount of XeF4 that could be produced is 0.5 moles.
XeF4 = Xe (g) 2 F2 (g) (g)
Xe and F2 have a mole ratio of 1:2. Because of this, the reaction would be limited by F2 when there is 1 mole of Xe and 1 mole of F2.
<h3>What is mole ratio?</h3>
The mole ratio is the ratio of any two compounds' mole amounts that are present in a balanced chemical reaction.
A comparison of the ratios of the molecules required to accomplish the reaction is given by the balancing chemical equation.
A mole ratio is a conversion factor used in chemical reactions to link the mole quantities of any two compounds. A conversion factor's numbers are derived from the balanced chemical equation's coefficients.
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Answer : Option C) The Octet Rule
Explanation : Atoms have a tendency to complete their outer energy level. This is known as Octet Rule.
The octet rule is a chemical rule of thumb which reflects the observation, that atoms of main-group elements tends to combine in such a way that each atom gets eight electrons in its valence shell, which gives it the same electron configuration as that of a noble gas.
In short, the tendency of an atom to fill its valence shell and attain a stable state it acquires or donates the electron is called as octet rule.
Answer:
51.79g Li₃P.
Explanation:
Li has a molar mass of 6.94 g (since there are 3, you multiply it 3 times) and P has a molar mass of 30.97 g. 6.94(3) + 30.97 = 51.79g.
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</h3>
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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Answer: Option (e) is the correct answer.
Explanation:
Formula to calculate radius is as follows.
p(r) = 
= 
= 0
+ 
= 0
= 
r = 
Thus, we can conclude that most likely radius at which the electron would be found is .
