Answer:
To have a high electrical conductivity, a normal metal should have mobile valence electrons and a long electron mean free path.
Explanation:
A long mean free path means that the electron goes a long distance between scattering events.
The chemical formula for this compound : Cu₂SO₄-Copper(I) sulfate
<h3>Further explanation</h3>
Given
2 atoms of Copper, 1 atom of Sulfur and 4 atoms of Oxygen.
Required
The chemical formula
Solution
The number of atoms forming a compound is usually indicated from the subscript in the formula
From the problem it can be seen that the compound has:
2 atoms of Cu, 1 atom of S, and 4 atoms of O
So the formula:
<em>Cu₂SO₄</em>
Answer:
pH = 2.
Explanation:
A weak acid is in equilibrium with its ions in a solution, so it must have an equilibrium constant (Ka). And, pKa = -logKa
Ka = 10⁻⁴
So, for CH₃COOH the equilibrium must be:
CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)
1 M 0 0 Initial
-x +x +x Reacted
1-x x x Equilibrium
And the equilibrium constant:
![Ka = \frac{[H+]x[CH3COO-]}{[CH3COOH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%2B%5Dx%5BCH3COO-%5D%7D%7B%5BCH3COOH%5D%7D)
Supposing x << 1:
10⁻⁴ = x²
x = √10⁻⁴
x = 10⁻² M, so the supposing is correct.
So,
pH = -log[H⁺]
pH = -log10⁻²
pH = 2
Answer:
1.23 g
Explanation:
<em>A chemist adds 1.55 L of a 0.00582 M calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sulfate the chemist has added to the flask.</em>
Step 1: Given data
- Volume of the solution (V): 1.55 L
- Molar concentration of the solution (C): 0.00582 M (0.00582 mol/L)
Step 2: Calculate the moles (n) of calcium sulfate added
We will use the following expression.
n = C × V
n = 0.00582 mol/L × 1.55 L
n = 0.00902 mol
Step 3: Calculate the mass corresponding to 0.00902 moles of calcium sulfate
The molar mass of calcium sulfate is 136.14 g/mol.
0.00902 mol × 136.14 g/mol = 1.23 g
