Answer:

Explanation:
a) Balanced equation
The balanced chemical equation for the titration is

b) pH at start
For simplicity, let's use B as the symbol for NH₃.
The equation for the equilibrium is

(i) Calculate [OH]⁻
We can use an ICE table to do the calculation.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.150 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.150 - x x x
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BBH%7D%5E%7B%2B%7D%5D%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.150%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:

(ii) Solve for x
![\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.150%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.150%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5Cx%5E%7B2%7D%20%3D%202.7%20%5Ctimes%2010%5E%7B-6%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.7%20%5Ctimes%2010%5E%7B-6%7D%7D%5C%5Cx%20%3D%20%5Ctext%7B%5BOH%5D%7D%5E%7B-%7D%20%3D%201.64%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctext%7B%20mol%2FL%7D)
(iii) Calculate the pH
(c) pH at equivalence point
(i) Calculate the moles of each species

B + HI ⇌ BH⁺ + I⁻
I/mol: 3.00 3.00 0
C/mol: -3.00 -3.00 +3.00
E/mol/: 0 0 3.00
(ii) Calculate the concentration of BH⁺
At the equivalence point we have a solution containing 3.00 mmol of NH₄I
Volume = 20.00 mL + 20.00 mL = 40.00 mL
![\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}](https://tex.z-dn.net/?f=%5Crm%20%5BBH%5E%7B%2B%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.00%20mmol%7D%7D%7B%5Ctext%7B40.00%20mL%7D%7D%20%3D%20%5Ctext%7B0.0750%20mol%2FL%7D)
(iii) Calculate the concentration of hydronium ion
We can use an ICE table to organize the calculations.
BH⁺+ H₂O ⇌ H₃O⁺ + B
I/mol·L⁻¹: 0.0750 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.0750 - x x x

![\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.0750%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-10%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.0750%20%5Ctimes%205.56%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%5E%7B2%7D%20%3D%204.17%20%5Ctimes%2010%5E%7B-11%7D%5C%5Cx%20%3D%20%5Csqrt%7B4.17%20%5Ctimes%2010%5E%7B-11%7D%7D%5C%5C%5Crm%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3Dx%20%3D%206.46%20%5Ctimes%2010%5E%7B-6%7D%5C%2C%20mol%20%5Ccdot%20L%5E%7B-1%7D)
(iv) Calculate the pH
![\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5Crm%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%3D%20-%5Clog%7B6.46%20%5Ctimes%2010%5E%7B-6%7D%7D%20%3D%20%5Clarge%20%5Cboxed%7B%5Cmathbf%7B5.19%7D%7D)
The titration curve below shows the pH at the beginning and at the equivalence point of the titration.
Answer:
Nuclear Chain Reactions. A chain reaction refers to a process in which neutrons released in fission produce an additional fission in at least one further nucleus. This nucleus in turn produces neutrons, and the process repeats. The process may be controlled (nuclear power) or uncontrolled (nuclear weapons).
Explanation:
M = 10.0 Kg
acceleration of gravity = 9.8 m/s²
F = M * a
F = 10.0 * 9.8
F = 98 newtons.
hope this helps!.
Because concentrated sulfuric acid is a strong oxidizer, so it has to be in water to react with metals and produce hydrogen properly. Sometimes dilute sulfuric acid is used, but that is another story