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3 years ago

What is evaporation? Describe what happens at the molecular level during evaporation.

1 answer:

6 0

Evaporation happens when atoms or molecules escape from the liquid and turn into a vapor. Not all of the molecules in a liquid have the same energy. Sometimes a liquid can be sitting in one place (maybe a puddle) and its molecules will become a gas. That's the process called evaporation. It can happen when liquids are cold or when they are warm. It happens more often with warmer liquids. You probably remember that when matter has a higher temperature, the molecules have a higher energy. When the energy in specific molecules reaches a certain level, they can have a phase change. Evaporation is all about the energy in individual molecules, not about the average energy of a system. The average energy can be low and the evaporation still continues.

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An isotope has 15 protons, 16 neutrons, and 15 electrons. What is the isotopic notation of the isotope

VikaD [51]

Answer:

$^{31}_{15}P$ is the isotopic notation of the atom

Explanation:

The isotope notation is:

$^a_bX$

Where a is the mass number = Number of protons + Number of neutrons

b is atomic number = Number of protons

The atomic number define the nature of the atom, the element with atomic number = 15 is phosphorus, P:

$^a_bP$

a = 15 protons + 16 neutrons = 31

b = 15

$^{31}_{15}P$ is the isotopic notation of the atom.

3 0

3 years ago

What would you have to do to the substance in

jenyasd209 [6]

By using thermal energy which is also heat energy the substance in cup A will melt becoming a liquid behaving like the substance in cup B

5 0

2 years ago

Role and significance of guassion or normal distribution?

kkurt [141]

Normal my guy!!!!!!!!!

5 0

3 years ago

Read 2 more answers

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?

Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0

2 years ago

Read 2 more answers