<span>C7H8
First, determine the number of relative moles of each element we have and the molar masses of the products.
atomic mass of carbon = 12.0107
atomic mass of hydrogen = 1.00794
atomic mass of oxygen = 15.999
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
We have 5.27 mg of CO2, so
5.27 / 44.0087 = 0.119749 milli moles of CO2
And we have 1.23 mg of H2O, so
1.23 / 18.01488 = 0.068277 milli moles of H2O
Since there's 1 carbon atom per CO2 molecule, we have
0.119749 milli moles of carbon.
Since there's 2 hydrogen atoms per H2O molecules, we have
2 * 0.068277 = 0.136554 milli moles of hydrogen atoms.
Now we need to find a simple integer ratio that's close to
0.119749 / 0.136554 = 0.876937
Looking at all fractions n/m where n ranges from 1 to 10 and m ranges from 1 to 10, I find a closest match at 7/8 = 0.875 with an error of only 0.001937, the next closest match has an error over 6 times larger. So let's go with the 7/8 ratio.
The numerator in the ratio was for carbon atoms, and the denominator was for hydrogen. So the empirical formula for toluene is C7H8.</span>
C.Work was required by an outside force.
Molarity is defined as the ratio of number of moles to the volume of solution in litres.
The mathematical expression is given as:
Here, molarity is equal to 1.43 M and volume is equal to 785 mL.
Convert mL into L
As, 1 mL = 0.001 L
Thus, volume = 
= 0.785 L
Rearrange the formula of molarity in terms of number of moles:
n = 
= 1.12255 mole
Now, Number of moles = 
Molar mass of potassium hydroxide = 56.10 g/mol
1.12255 mole = 
mass in g = 
= 62.97 g
Hence, mass of 
= 62.97 g
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
