Answer:
D. Scientific notation is a way of writing large and small numbers
Answer:
0.504 M
Explanation:
Step 1: Write the balanced neutralization reaction
2 KOH + H₂SO₄ ⇒ K₂SO₄ + 2 H₂O
Step 2: Calculate the reacting moles of KOH
55.2 mL (0.0552 L) of 0.500 M KOH react. The reacting moles of KOH are:
0.0552 L × 0.500 mol/L = 0.0276 mol
Step 3: Calculate the moles of H₂SO₄ that reacted with 0.0276 moles of KOH
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of H₂SO₄ are 1/2 × 0.0276 mol = 0.0138 mol
Step 4: Calculate the concentration of H₂SO₄
0.0138 moles of H₂SO₄ are in 27.4 mL (0.0274 L). The molarity of H₂SO₄ is:
[H₂SO₄] = 0.0138 mol/0.0274 L = 0.504 M
Answer:
AgI, AgBr, AgCl and Ag₂CrO₄
Explanation:
Ksp (product solubility constant) is defined as the equilibrium constant of the general reaction:
XₐYₙ(s) → aXⁿ⁺(aq) + nYᵃ⁻(aq)
<em>Where X is cation and Y is anion.</em>
Ksp = [aXⁿ⁺]ᵃ [nYᵃ⁻]ⁿ
The presence of XₐYₙ(s) produce ax moles of aXⁿ⁺ and nx moles of Yᵃ⁻. <em>Where X is the solubility of the compound.</em>
Replacing in Ksp:
Ksp = [ax]ᵃ [nx]ⁿ
Solving for x, Solubility (S) is defined as:
![S = \sqrt[n+a]{\frac{Ksp}{a^{a} n^n} }](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%5Bn%2Ba%5D%7B%5Cfrac%7BKsp%7D%7Ba%5E%7Ba%7D%20n%5En%7D%20%7D)
For AgCl, Ag₂CrO₄, AgBr and AgI solubilities are:
![S = \sqrt[2]{\frac{1.8x10^{-10}}{1} }](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%5B2%5D%7B%5Cfrac%7B1.8x10%5E%7B-10%7D%7D%7B1%7D%20%7D)
= 1.34x10⁻⁵M
![S = \sqrt[3]{\frac{1.1x10^{-12}}{4} }](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B1.1x10%5E%7B-12%7D%7D%7B4%7D%20%7D)
= 6.50x10⁻⁵M
![S = \sqrt[2]{\frac{5.4x10^{-13}}{1} }](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%5B2%5D%7B%5Cfrac%7B5.4x10%5E%7B-13%7D%7D%7B1%7D%20%7D)
= 7.35x10⁻⁷M
![S = \sqrt[2]{\frac{8.5^{-17}}{1} }](https://tex.z-dn.net/?f=S%20%3D%20%5Csqrt%5B2%5D%7B%5Cfrac%7B8.5%5E%7B-17%7D%7D%7B1%7D%20%7D)
= 9.22x10⁻⁹M
The lower solubility is the first compound in precipitate, thus, order of precipitation is:
<em>AgI, AgBr, AgCl and Ag₂CrO₄</em>
<h3><u>Answer</u>;</h3>
= 226 Liters of oxygen
<h3><u>Explanation</u>;</h3>
We use the equation;
LiClO4 (s) → 2O2 (g) + LiCl, to get the moles of oxygen;
Moles of LiClO4;
(500 g LiClO4) / (106.3916 g LiClO4/mol)
= 4.6996 moles
Moles of oxygen;
But, for every 1 mol LiClO4, two moles of O2 are produced;
= 9.3992 moles of Oxygen
V = nRT / P
= (9.3992 mol) x (8.3144621 L kPa/K mol) x (21 + 273) K / (101.5 kPa)
= 226 L of oxygen
