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Butoxors [25]
2 years ago
13

Select the correct answer from each drop-down menu.

Chemistry
2 answers:
kirza4 [7]2 years ago
7 0

The medium sized puddle of spilled hot water has the highest vapor pressure because vapor pressure depends on temperature of the object.

Answer: B

Explanation

Vapor pressure, the pressure of vapor molecules of liquid exerting on the liquid surface.

It helps to determine the evaporation rate, the more the vapor pressure the more rate of evaporation will occur.

This is because the vapor pressure is the quantification of vapor molecules of the liquid on the surface of the liquid on giving an external force like heat.

The vapor pressure depends on liquid temperature and liquid surface area and pressure.

So in the given case, rainwater puddle has small surface area and room temperature, spilled hot water puddle has higher surface area as well as temperature compared to rain water puddle.

Even though the cold water puddle has large surface area but the temperature of cold water is very less compared to the temperature of hot water, so it makes the medium sized hot water puddle the one with the highest vapor pressure as the vapor pressure depends on temperature of the liquid.

tankabanditka [31]2 years ago
3 0

Answer:

The medium-sized puddle of spilled hot water has the highest vapor pressure

Explanation:

The evaporation of water is a clear example of phase change from liquid to gas. The chemicals potentials for both phases α (liquid) and β (vapor) are functions dependig of the temperature (T) and the pressure (P) and they have the same value.

μα(T,P) = μβ(T,P)

Based in that assumption is that the equation of Clausius Clapeyron:

Ln Pv = - L/(RT) +C

Being Pv, the vapor pressure. L the vapor entalpy and R the gas constant.

So, the vapor pressure is a mere function of the temperature, thus if the temperature rises, the vapor pressure increases.

With that said, the puddle of hot water, has more temperature and thus, a higher vapor pressure.

Hope it helped!

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A piston containing 0.120moles of methane gas, CH4, has a volume of 2.12liters. If methane is added until the volume is increase
earnstyle [38]

Answer:

2.83 g

Explanation:

At constant temperature and pressure, Using Avogadro's law

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₁ = 2.12 L

V₂ = 3.12 L

n₁ = 0.120 moles

n₂ = ?

Using above equation as:

\frac{2.12}{0.120}=\frac{3.12}{n_2}

2.12n_2=0.12\cdot \:3.12

2.12n_2=0.3744

n_2=\frac{0.3744}{2.12}

n_2=0.17660

n₂ = 0.17660 moles

Molar mass of methane gas = 16.05 g/mol

So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g

<u>2.83 g  are in the piston.</u>

8 0
2 years ago
For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati
AveGali [126]

Answer:

a. withdraws electrons inductively

b. donates electrons by hyperconjugation

c. donates electrons by resonance

d.  withdraws electrons inductively

Explanation:

a.  The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:

F > Cl > Br>I

b.  The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons.  Because of the delocalization, hyperconjugation is the result effect.

c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

3 0
2 years ago
The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his
ziro4ka [17]

Answer:

The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group

The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists

Explanation:

5 0
2 years ago
Elements that’s are gases, brittle &amp; poor conductors at room temperature are
vlabodo [156]

I think the answer is C for this question

5 0
2 years ago
Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis
almond37 [142]

Answer: a) BaCl_2

b)  BiO_3H_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = \frac{0.48}{0.48}=1

For O =\frac{0.96}{0.48}=2

The ratio of Ba: Cl= 1:2

Hence the empirical formula is BaCl_2

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles

Moles of O= \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= \frac{0.38}{0.38}=1

For O =\frac{1.15}{0.38}=3

For H=\frac{1.16}{0.38}=3

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is BiO_3H_3

6 0
2 years ago
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