Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.
Amount of heat required to vaporize 1 mole of lead = 177.7 kJ
Molar mass of lead = 207.2 g
Mass of lead given = 1.31 kg = 1310 g (1kg=1000g)
Heat required to vaporize 207.2 of lead = 177.7 kJ
Thus Heat required to vaporize 1310 g of lead =
Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point
The answer is A. How does the cuddle fish change its colors ?
Answer:
b. 50ppm of Pb
c. 2x10⁻⁴M
j. no
Explanation:
The diluted sample has a concentration of 0.5 ppm of Pb. The sample was diluted from 1mL to 100mL. That means the dilution factor is:
100mL / 1mL = 100
That means, the sample was diluted 100 times from its original concentration.
As diluted sample has a concentration of 0.5ppm, the undiluted sample has a concentration of
0.5ppm×100 = <em>50ppm of Pb</em>
1 ppm means 1x10⁻⁴ % by mass, 50 ppm are 50x10⁻⁴ % by mass
As molar mass of Pb is 207g/mol, molarity of 0.050g / L a(50 ppm) are:
0.050gₓ (1mol / 207g) / 1L = <em>2x10⁻⁴M</em>
<em></em>
Based on EPA regulation, the maximum concentration of Lead in drinking water must be 0.015ppm. That means this water <em>is not safe to drink</em>
12.0g x 1 mol / 63.546g = 0.188839581mol
<span>So, for every 1 mole, we have 6.022 x 10^23 of whatever we're measuring. This gives us a conversion factor of (1 mole / 6.022 x 10^23 atoms) or (6.022 x 10^23 atoms / 1 mole).
</span>
0.188839581 mol x (6.022 x 10^23 atoms) / 1 mol = 1.137191955 x 10^23
<span>Remember from before that we are limited to 3 significant figures. Since our calculations are complete, we can now round down to: 1.14 x 10^23 </span>
<span>That should be your answer!
Hope it helps!
xo</span>
