Answer:
Electrolysis, process by which electric current is passed through a substance to effect a chemical change. ... Electric current (i.e., electrons) enters through the negatively charged electrode (cathode); components of the solution travel to this electrode, combine with the electrons, and are transformed (reduced)
Answer:
Explanation:
Given that;
diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m
length (l) = 10 cm = 0.1 m
porosity = 50%
height (h) = 30 cm = 0.3 m
time (t) = 5 s
volume (v) = 60 cm³ = 60 × 10⁻⁶ m³
Q (flow rate) = 
Q = 
Q = 
From constant head method, we use the relation K = 
to determine the hydraulic conductivity ; we have:
Seepage velocity 
where; velocity = 
=
= 
Answer:
59.077 kJ/mol.
Explanation:
- From Arrhenius law: <em>K = Ae(-Ea/RT)</em>
where, K is the rate constant of the reaction.
A is the Arrhenius factor.
Ea is the activation energy.
R is the general gas constant.
T is the temperature.
- At different temperatures:
<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>
k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.
ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]
∴ ln(3) = 1.859 x 10⁻⁵ Ea
∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.
Answer:
The answer to your question is maybe letter D, but the last oxygen needs a number 6.
Explanation:
The empirical formula gives the actual elements that form part of a molecule but not the total numbers.
The molecular formula gives the total number of atoms of each element in a molecule.
We must factor the molecular formula to know if a formula is the empirical formula of that.
A. CH₄ C₂H₆ = 2(CH₃) these are not empirical molecular formulas
B. CH₂O C₄H₆O these are not empirical-molecular formulas
C. O₂ O₃ these are not empirical-molecular formulas
D. C₃H₄O₃ C₆H₈O these are not empirical-molecular formulas
the last oxygen needs a number 6 to be
the answer.
<h3>
Answer:</h3>
2.04 mol CBr₄
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Organic</u>
- Writing Organic Compounds
- Writing Covalent Compounds
- Organic Prefixes
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
675 g CBr₄
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of Br - 79.90 g/mol
Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol
<u>Step 3: Convert</u>
<u />
<u />
<u />
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄
