Answer: option A. 2
Explanation: in the formula Sr3(PO4)2, the 2 behind (PO4) is affecting both P and O4. It means that we have P2 in the formula
Move anywhere and can be anywhere while it is in that phase
<span>So to make it clear let's break the equation down species by species and assess the number of each species on bothe sides of the equation:
2C</span>₈H₈ + 25O₂ → 8CO₂ + 18H₂<span>O
LHS: C - 16 RHS: C - 8
H - 16 H - 36
O - 50 O - 34
Thus based on that it is evident that the equation is not quite balanced. This therefore means a "</span><span>No, because the number of carbon, hydrogen & oxygen atoms on both sides of the equation are not equal."
</span>The actual balance equation would be C₈H₈ + 10O₂ → 8CO₂ + 4H₂O
Answer:
2.97 × 10¹³ g
Explanation:
First, we have to calculate the biomass the is burned. We can establish the following relations:
- 2.47 acre = 10,000 m²
- 10 kg of C occupy an area of 1 m²
- 50% of the biomass is burned
The biomass burned in the site of 400,000 acre is:

Let's consider the combustion of carbon.
C(s) + O₂(g) ⇒ CO₂(g)
We can establish the following relations:
- The molar mass of C is 12.01 g/mol
- 1 mole of C produces 1 mole of CO₂
- The molar mass of CO₂ is 44.01 g/mol
The mass of produced is CO₂:

#2. Percent error (or deviation) is calculated as ((TV-EV)/EV))*100