HCl04 is a strong acid solution. So, in theory everything dissociates.
pH= -log(0.000081)= 4.09
Answer:
1552.83J Released
Explanation:
1. mass/m=225
Initial temp:86C, final:32.5C
Changed Temp: 32.5-86= -53.5C
s=0.129 J/gC
Formula: q= m times s times changed Temp.
q=(225)(0.129)(-53.5)
q= -1552.83 J
q=1552.83 J Released
The partial pressure of oxygen, P(O₂) is 198.83 mmHg.
<h3>What is partial pressure of a gas?</h3>
The partial pressure of a gas is the pressure gas will exert in a mixture of gases which do not react chemically together.
The sum of partial pressure of gases at atmospheric pressure = 760 mmHg
P(N₂) + P(H₂O) + P(CO₂) + P(O₂) + P(other) = 760 mmHg
P(N₂) = 0.72 atm = 547.2 mmHg
P(H₂O) = 7.7 torr = 7.7 mmHg
P(CO₂) = 0.37 mmHg
P(other) = 0.97 kPa = 5.9 mmHg
P(O₂) = 760 - (547.2 + 7.7 + 0.37 + 5.9) = 198.83 mmHg
Therefore, the partial pressure of oxygen, P(O₂) is 198.83 mmHg.
Learn more partial pressure at: brainly.com/question/14119417
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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:
Esta configuración electrónica pertenece al elemento, neón.
Si tienes preguntas, házmelo saber
. :)