Answer:
A its a step by step equation the answer is A
Answer:
Do research on a Particular topic
Answer:
<h3>601.93 g/mol</h3>
<h3>explanation:</h3>
Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.
Answer:

Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 58.12 44.01
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
m/g: 9.511
1. Moles of C₄H₁₀

2. Moles of CO₂
The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

3. Mass of CO₂
