CH3NH2 can only have as many hydrogen bonds as hydrogen bonding sites in the molecule. CH3NH2 has two N−H bonds and a lone pair of electrons on the nitrogen atom. Therefore, CH3NH2 can form three hydrogen bonds with water.
![n(\text{C}_3\text{H}_7\text{OH}) = N(\text{C}_3\text{H}_7\text{OH}) / N_A\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 5.2 \times 10^{21} / (6.02 \times 10^{23})\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 8.6 \times 10^{-3} \; \text{mol}](https://tex.z-dn.net/?f=n%28%5Ctext%7BC%7D_3%5Ctext%7BH%7D_7%5Ctext%7BOH%7D%29%20%3D%20N%28%5Ctext%7BC%7D_3%5Ctext%7BH%7D_7%5Ctext%7BOH%7D%29%20%2F%20N_A%5C%5C%20%5Cphantom%7Bn%28%5Ctext%7BC%7D_3%5Ctext%7BH%7D_7%5Ctext%7BOH%7D%29%7D%20%3D%205.2%20%5Ctimes%2010%5E%7B21%7D%20%2F%20%286.02%20%5Ctimes%2010%5E%7B23%7D%29%5C%5C%20%5Cphantom%7Bn%28%5Ctext%7BC%7D_3%5Ctext%7BH%7D_7%5Ctext%7BOH%7D%29%7D%20%3D%208.6%20%5Ctimes%2010%5E%7B-3%7D%20%5C%3B%20%5Ctext%7Bmol%7D)
where
the Avogadro's constant that relates the number of particles to their number, in the unit moles
.
The molar mass of propanol- mass per mole propanol- can be directly deduced from its molecular formula with reference to a modern periodic table.
![M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}](https://tex.z-dn.net/?f=M%28%5Ctext%7BC%7D_3%5Ctext%7BH%7D_7%5Ctext%7BOH%7D%29%20%3D%20%5Cunderbrace%7B3%20%5Ctimes%2012.01%7D_%7B%5Ctext%7Bcarbon%7D%7D%20%2B%20%5Cunderbrace%7B8%20%5Ctimes%201.008%7D_%7B%5Ctext%7Bhydrogen%7D%7D%20%2B%20%5Cunderbrace%7B1%5Ctimes%2016.00%7D_%7B%5Ctext%7Boxygen%7D%7D%20%3D%2060.09%20%5C%3B%20%5Ctext%7Bg%7D%20%5Ccdot%20%5Ctext%7Bmol%7D%5E%7B-1%7D)
of propanol molecules would thus have a mass of ![8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}](https://tex.z-dn.net/?f=8.6%20%5Ctimes%2010%5E%7B-3%7D%20%5C%3B%20%5Ctext%7Bmol%7D%20%5Ctimes%2060.09%20%5C%3B%20%5Ctext%7Bg%7D%20%5Ccdot%20%5Ctext%7Bmol%7D%5E%7B-1%7D%20%3D%200.52%20%5C%3B%20%5Ctext%7Bg%7D)
Answer:
B - What we change
Explanation:
Dependent Variable - What we measure
Control Variable - what stays the same
Conclusion - what we conclude
<em>Hope</em><em> </em><em>this</em><em> </em><em>can</em><em> </em><em>Help</em><em>!</em>
<em>:</em><em>D</em>
Answer:
Q = 114349.5 J
Explanation:
Hello there!
In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:
![Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J](https://tex.z-dn.net/?f=Q_1%3D45g%2A4.18%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%2A%28100%5C%C2%B0C-40%5C%C2%B0C%29%3D11286J%5C%5C%5C%5CQ_2%3D45g%2A%202260%20%5Cfrac%7BJ%7D%7Bg%7D%20%3D101700J%5C%5C%5C%5CQ_3%3D45%2A2.02%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%2A%28115%5C%C2%B0C-100%5C%C2%B0C%29%3D1363.5J)
Thus, the total energy turns out to be:
![Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J](https://tex.z-dn.net/?f=Q_T%3D11286J%2B101700J%2B1363.5J%5C%5C%5C%5CQ_T%3D114349.5J)
Best regards!