All of the above are correct answers
C6H15, C2H5 has a molar mass of 29g/mol. 87 divided by 29 is 3. Then multiply each element subscript by 3
a. pH=2.07
b. pH=3
c. pH=8
<h3>Further explanation</h3>
pH=-log [H⁺]
a) 0.1 M HF Ka = 7.2 x 10⁻⁴
HF= weak acid
![\tt [H^+]=\sqrt{Ka.M}\\\\(H^+]=\sqrt{7.2.10^{-4}\times 0.1}\\\\(H^+]=8.5\times 10^{-3}\\\\pH=3-log~8.5=2.07](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D%5Csqrt%7B7.2.10%5E%7B-4%7D%5Ctimes%200.1%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D8.5%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~8.5%3D2.07)
b) 1 x 10⁻³ M HNO₃
HNO₃ = strong acid
![\tt pH=-log[1\times 10^{-3}]=3](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-3%7D%5D%3D3)
c) 1 x 10⁻⁸ M HCl
![\tt pH=-log[1\times 10^{-8}]=8](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1%5Ctimes%2010%5E%7B-8%7D%5D%3D8)
Answer:
Option D. 3, 1, 3, 1
Explanation:
From the question given above,
HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
The equation can be balance as follow:
HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
There are 3 atoms of N on the right side and 1 atom on the left side. It can be balance by 3 in front of HNO₃ as shown below:
3HNO₃ + Al(OH)₃ —> HOH + Al(NO₃)₃
There are a total of 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by 3 in front of HOH as shown below:
3HNO₃ + Al(OH)₃ —> 3HOH + Al(NO₃)₃
Now, the equation is balanced.
Thus, the coefficients are 3, 1, 3, 1
