Answer:
Explanation:
1. Concentration of SO₄²⁻
SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷
0.0150 x
![K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%200.0150x%20%3D%203.44%20%5Ctimes%2010%5E%7B-7%7D%5C%5Cx%20%3D%20%5Cdfrac%7B3.44%20%5Ctimes%2010%5E%7B-7%7D%7D%7B0.0150%7D%20%3D%20%5Cmathbf%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Concentration of Pb²⁺
PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸
x 2.293 × 10⁻⁵
![K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%20x%20%5Ctimes%202.293%20%5Ctimes%2010%5E%7B-5%7D%20%3D%202.53%20%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5Cx%20%3D%20%5Cdfrac%7B2.53%20%5Ctimes%2010%5E%7B-8%7D%7D%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%20%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20concentration%20of%20Pb%24%5E%7B2%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
HCl Acid + Sodium Hydroxide ----> Sodium Chloride + water.
<u>Explanation</u>:
- The reaction between an acid and a base is known as a neutralization reaction. The reaction of an acid with a base to give salt, water and heat is called neutralization.
- When hydrochloric acid reacts with sodium hydroxide, sodium chloride and water are produced.
HCl + NaOH → NaCl + H2O + Heat
- The heat evolved in the neutralization reaction raises the temperature of the reaction mixture.
- An electro-electrodialysis process (EED) is utilized to create HCl and NaOH from exchange NaCl. NaOH and HCl arrangements with purity higher than 99.9% are acquired. The experimental estimations of the transitions for HCl and NaOH are contrasted and values determined from the incorporation of the Nernst–Planck electro dispersion conditions.
The corect answers will be:
1) A
2) B
3) D
Hope this helped :)
NH4I (aq) + KOH (aq) in chemical equation gives
NH4I (aq) + KOH (aq) = KI (aq) + H2O(l) + NH3 (l)
Ki is in aqueous state H2o is in liquid state while NH3 is in liquid state
from the equation above 1 mole of NH4I (aq) react with 1 mole of KOH(aq) to form 1mole of KI(aq) , 1mole of H2O(l) and 1 Mole of NH3(l)
d. When aluminum-28 undergoes beta decay, silicon-28 is produced.
Explanation:
When the nuclei of aluminium-28 decays, it produces silicon- 28:
Aluminium ²⁸₁₃Al
Silicon 28 ²⁸₁₄Si
beta particle ⁰₋₁
²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁
This way, the mass and atomic number are conserved.
Conservation of mass number:
28 = 28 + 0, 28 = 28
13 = 14 -1 , 13 = 13
Learn more:
Balancing nuclear equations brainly.com/question/10094982
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