Answer : The reaction requires 242 grams of HF.
Explanation : Given,
Mass of
= 182 g
Molar mass of
= 60 g/mole
Molar mass of HF = 20 g/mole
First we have to calculate the moles of ![SiO_2](https://tex.z-dn.net/?f=SiO_2)
![\text{Moles of }SiO_2=\frac{\text{Mass of }SiO_2}{\text{Molar mass of }SiO_2}=\frac{182g}{60g/mole}=3.03moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DSiO_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DSiO_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DSiO_2%7D%3D%5Cfrac%7B182g%7D%7B60g%2Fmole%7D%3D3.03moles)
Now we have to calculate the moles of HF.
The given balanced chemical reaction is,
![SiO_2+4HF\rightarrow SiF_4+2H_2O](https://tex.z-dn.net/?f=SiO_2%2B4HF%5Crightarrow%20SiF_4%2B2H_2O)
From the balanced chemical reaction, we conclude that
As, 1 mole of
react with 4 moles of HF
So, 3.03 moles of
react with
moles of HF
Now we have to calculate the mass of HF.
![\text{Mass of HF}=\text{Moles of HF}\times \text{Molar mass of HF}=12.12mole\times 20g/mole=242g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20HF%7D%3D%5Ctext%7BMoles%20of%20HF%7D%5Ctimes%20%5Ctext%7BMolar%20mass%20of%20HF%7D%3D12.12mole%5Ctimes%2020g%2Fmole%3D242g)
Therefore, the reaction requires 242 grams of HF.
Answer:
5.7
Explanation:
The pH can be calculated by:
pH = -log[H⁺]
Where [H⁺] is the concentration of hydrogen ions. The pH is a way to identify if the solution is an acid or a base. When pH is equal to 7, the solution is neutral, when it is less than 7, it is acid, and when it is larger than 7, it's a base.
So, for the acid solution given, [H⁺] = 2.0x10⁻⁶ M
pH = -log(2.0x10⁻⁶)
pH = 5.7
Answer:
Kb = ![4.45\times10^{-7}\ mol/L](https://tex.z-dn.net/?f=4.45%5Ctimes10%5E%7B-7%7D%5C%20mol%2FL)
![p^{Kb}=6.35](https://tex.z-dn.net/?f=p%5E%7BKb%7D%3D6.35)
Explanation:
For a weak organic base, the formula to find
is given by:
![p^{OH}=p^{K_b}+\log c](https://tex.z-dn.net/?f=p%5E%7BOH%7D%3Dp%5E%7BK_b%7D%2B%5Clog%20c)
where c is the concentration of base.
Here c= ![5\times10^{-3}\ M](https://tex.z-dn.net/?f=5%5Ctimes10%5E%7B-3%7D%5C%20M)
![p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05](https://tex.z-dn.net/?f=p%5E%7BH%7D%3D9.95%5C%5Cp%5E%7BOH%7D%3D14-p%5E%7BH%7D%3D14-9.95%3D4.05)
Substituting the above values in the formula,we get:
![p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L](https://tex.z-dn.net/?f=p%5E%7Bk_b%7D%3Dp%5E%7BOH%7D-%5Clog%20c%5C%5Cp%5E%7Bk_b%7D%3D4.05-%5Clog%20%285%5Ctimes10%5E%7B-3%7D%29%5C%5Cp%5E%7BK_b%7D%3D6.35%5C%5CK_b%3D%24antilog%206.35%3D4.45%5Ctimes10%5E%7B-7%7D%5C%20mol%2FL)
Hence:
Kb = ![4.45\times10^{-7}\ mol/L](https://tex.z-dn.net/?f=4.45%5Ctimes10%5E%7B-7%7D%5C%20mol%2FL)
![p^{Kb}=6.35](https://tex.z-dn.net/?f=p%5E%7BKb%7D%3D6.35)