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miss Akunina [59]
3 years ago
14

Why does boiling water in a large pot on a stove have a temperature of 100°C even if it has been boiling for an hour?

Chemistry
2 answers:
Pani-rosa [81]3 years ago
5 0

Answer:

The heat energy that is added to the water is used to vaporize the water rather than raise its temperature.

Explanation:

The boiling point of water is 100 °C, which means that when water reaches this temperature, it will vaporize rather than rise in temperature. Under normal conditions, the temperature of water cannot rise above 100 °C.

Arisa [49]3 years ago
3 0
Because 100 degrees is the boiling point, the water boils and becomes steam at that point. So the water that was higher than 100 degrees actually became steam, which is why the water in the pot is tops off at 100 degrees.
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The periodic table contains information about which types of matter?
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If the atomic mass of 1 oxygen atom is 15.9994 amu, how much mass does of 1 mole oxygen have?​
IrinaK [193]

Answer:

15.9994 amu

Explanation:

By definition 1 mole of any substance is the mass that contains 1 Avogadro's Number of particle. (=> 6.02 x 10²³ particles/mole).

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1 mole Sulfur => 32.064 amu => contains 6.02 x 10²³ particles Sulfur /mole

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3 years ago
What main criterion determines major divisions in the geologic time scale?
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4 0
3 years ago
You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho
timama [110]

Explanation:

As per the given data, at a higher temperature, at 24^{o}C, the solution will occupy a larger volume than at 20^{o}C.

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

                 = \frac{C_{1}}{d_{1}}

                 = \frac{1.000 mol/L}{0.9982071 g/ml}

                 = 1.0017961 mol/g  

At 24^{o}C, density of water = 0.9972995 g/ml

Since, \frac{concentration}{density} = \frac{C_{2}}{d_{2}}

                           = \frac{C_{2}}{0.9972995}

Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

so,                   1.0017961 mol/g = \frac{C_{2}}{0.9972995}

                      C_{2} = 1.0017961 \times 0.9972995

                                  = 0.9990907 mol/L

Therefore, in 500 ml, concentration of KNO_{3} present is calculated as follows.

             C_{2} = \frac{concentration}{volume}

               0.9990907 mol/L = \frac{concentration}{0.5 L}  

               concentration = 0.49954537 mol

Hence, mass (m'') = 0.49954537 mol \times 101.1032 g/mol = 50.5056 g       (as molar mass of KNO_{3} = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

                      m = m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}

where,          d_{air} = density of air = 0.0012 g/ml

                     d_{weight} = density of callibration weights = 8.0g/ml                      

                     d = density of weighed object

Hence, the true mass will be calculated as follows.

           True mass(m) = 50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}

             true mass(m) = 50.5268 g

                                  = 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of KNO_{3} needs to be measured.

8 0
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