For the purpose of proper representation in this item, we let the number of moles of carbon in the compound be x, that of H is y. The equation of toluene now becomes,
CxHy
The combustion reaction is,
CxHy + O2 --> CO2 + H2O
The equation presented above may not be balanced yet. Then, we determine the number of mmols of C, H, and O in the product using the given masses.
(1) 8.20 mg CO2
(8.2 mg CO2)(1 mmol CO2/44 mg CO2) = 0.186 mmol CO2
which means,
0.186 mmol C
0.373 mmol O
(2) 1.92 mg H2O
(1.92 mg H2O)(1 mmol H2O/18 mg H2O) = 0.107 mmol H2O
which means
0.2133 mmol H
0.107 mmol O
Thus, the equation for toluene is,
C(0.186)H(0.2133)
Dividing the numbers by the lesser value,
CH(8/7)
To eliminate the fraction, we multiply by the denominator. Thus, the final answer would be,
<em> C7H8</em>
Answer a:
<span>A 1.5 V battery, the electromagnet picked up an average of 6 paper clips, while with the 6.0 V battery, an average of 23 paper clips were picked up. Battery of 6.0V is 6.0/1.5 = 4 times stronger than battery of 1.5 V
Answer b:
</span><span>Ratio of the number of paper clips picked up using the 6.0 V battery to the number picked up using the 1.5 V battery is = 23/6 = 3.8 </span>≈ 4.
Answer c:
As the voltage power increase, more paper clips were picked up by electromagnet. This indicated that there is a direct relationship. Mathematically it can be expressed as:
Voltage Power α Number of paper clips that were picked up
According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.
V∝ T, P is constant
where V, T and P are volume, temperature and pressure
= 
where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.
Answer:
empirical formula = C3H7
molecular formula = C6H14
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g