2Al₂O₃ = 4Al + 3O₂
M(A₂O₃)=101.96 g/mol
m(Al₂O₃)=250 g
n(O₂)=3m(Al₂O₃)/{2M(Al₂O₃)}
n(O₂)=3*250/{2*101.96}=3.678 mol
There are 20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.
<h3>How many molecules in 3.4 moles of NH4NO3?</h3>
We know that one mole of a substance has 6.022 × 10²³ molecules so in 3.4 moles of NH4NO3, we have 20.5 x 10^24 molecules if we multiply the 6.022 × 10²³ with 3.4.
So we can conclude that there are 20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.
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<span>The chemical formula is pretty straightforward. 2KOH reacts to produce H2O and K2O. This is the balanced chemical reaction between: Solid potassium hydroxide koh decomposing into gaseous water and solid potassium.</span>
I just took that quiz , it’s A
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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