════════ ∘◦❁◦∘ ════════
<h3>Answer = 2</h3>
════════════════════
<h3>Known</h3>
Mass = 32grams
Name of atom = Methane (CH4)
Molar mass C = 12
Molar mass H = 1
════════════════════
<h3>Question</h3>
molecules (mol?)
════════════════════
<h3>Way to do</h3>
mass = mol × molar mass
32g = mol × (CH4)
33g = mol × (12 + (4×1))
32g = mol × 16
mol = 32 : 16
mol = 2
════════════════════
#Ey tell me if its mol or total particles
Answer:
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Explanation:
Step 1: The balanced equation
AB2 ⇒ A2+ + 2B-
Step 2: Given data
Concentration of A2+ = 0.00253 M
Concentration of B- = 0.00506 M
Step 3: Calculate the equilibrium constant
Equilibrium constant Ksp of [AB2] = [A2+][B-]²
Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Answer:
Mass of Rb-87 is 86.913 amu.
Explanation:
Given data:
Average mass of rubidium = 85.4678 amu
Mass of Rb-85 = 84.9117
Ratio of 85Rb/87Rb in natural rubidium = 2.591
Mass of Rb = ?
Solution:
The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.
For 100% naturally occurring Rb = 2.591 + 1 = 3.591
% abundance of Rb-85 = 2.591/ 3.591 = 0.722
% abundance of Rb-87 = 1 - 0.722= 0.278
84.9117 × 0.722 + X × 0.278 = 85.4678
61.306 + X × 0.278 = 85.4678
X × 0.278 = 85.4678 - 61.306
X × 0.278 = 24.1618
X = 24.1618 / 0.278
X = 86.913 amu
The correct answer is A. Can I get the Brainliest?
