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AveGali [126]
3 years ago
9

WILL AWARD BRAINLIEST! I NEED HELP PLEASE! Match the states of matter for each of the five lines below.

Chemistry
2 answers:
KonstantinChe [14]3 years ago
5 0

Answer:

see explanations

Explanation:

The graphic is the heating curve for water. Note that it is divided into 5 distinct heat flow segments. The segments with changing slopes are single phase segments with changes in temperature values. From left to right segment A is solid ice being warmed to it's melting point. Segment B is the melting segment in which 2 phases are in contact (solid + liquid). Note that addition of heat does not change the temperature. Segment C is warming of the liquid (single phase) up to its boiling point. At the boiling point the liquid begins to pass into the gas phase and again 2 phases are in contact; i.e., liquid & gas. Note again when two phases are in contact no temperature change occurs. Finally, segment E is the heating of the pure, single phase gas.

In summary ...

Segment A => heating single phase (solid) ice up to melting pt.

Segment B => melting of ice => 2 phases in contact (s & l) ΔT = ∅.

Segment C => heating single phase (liquid) water up to boiling pt.

Segment D => boiling of liquid => 2 phases in contact (l & g). ΔT = ∅.

Segment E => heating single phase (steam) up to desired temperature.

For what it's worth, the equation for the segments that show increasing temperature values is q = mcΔT (m= mass, c = specific heat & ΔT temp change.

The segments with zero slopes (horizontal lines) are defined by equations  q = m·ΔHₓ where m = mass & ΔHₓ = heat of fusion (a constant = 335 j/g). The same is true for the line at 100°C where q = m·ΔH(v) where m = mass & ΔH(v) is the heat of vaporization (a constant = 2259 j/g.

Calculations involve calculating the amount heat transfer for each segment individually and then adding the heat values to obtain the total heat transfer.

If you need more instruction on this topic, kick back a note and I'll try to help clarify. Good Luck, Doc :-)

damaskus [11]3 years ago
5 0

Answer:

Picture attached has answers

Explanation:

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Determine the a) energy (in eV) and b) wavelength (in cm) corresponding to blue light of frequency 670 THz
Lesechka [4]

Answer :

(a) The energy of blue light (in eV) is 2.77 eV

(b) The wavelength of blue light is 4\times 10^{-5}cm

Explanation:

The relation between the energy and frequency is:  

Energy=h\times Frequency

where,

h = Plank's constant = 6.626\times 10^{-34}J.s

Given :

Frequency = 670THz=670\times 10^{12}s^{-1}

Conversion used :

1THz=10^{12}Hz\\1Hz=1s^{-1}\\1THz=10^{12}s^{-1}

So,  

Energy=(6.626\times 10^{-34}J.s)\times (670\times 10^{12}s^{-1})

Energy=4.44\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(4.44\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=2.77eV

The energy of blue light (in eV) is 2.77 eV

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where,

c = the speed of light = 3\times 10^8m/s

Frequency = 670\times 10^{12}s^{-1}

So, Wavelength is:

670\times 10^{12}s^{-1}=\frac{3\times 10^8m/s}{Wavelength}

Wavelength=\frac{3\times 10^8m/s}{670\times 10^{12}s^{-1}}=4\times 10^{-7}m=4\times 10^{-5}cm

Conversion used : 1m=100cm

The wavelength of blue light is 4\times 10^{-5}cm

7 0
3 years ago
Are CH4 and CH3OH soluble in water?
Serhud [2]

CH4 is <u>not</u> soluble in water

whereas CH3OH <u>is</u> soluble in water.

7 0
3 years ago
Read 2 more answers
What accounts for the majority of all air pollution? Is that source increasing or decreasing overall?
lianna [129]
Hi there!

One of the main causes of air pollution is smoking.

Smoking affects not only humans, but pets and animals as well. Animals can get cancer too!

Another big cause of pollution is cars. Gasoline of course comes out into the air after it is used. It also affects animals.

A lot of people know that we should change this, but hardly anyone is really doing anything to try and stop it.

I know this is a short answer, but I hope it helps!
4 0
3 years ago
Read 2 more answers
A sample of gas in a balloon has an initial temperature of 21 ∘C and a volume of 1310 L . If the temperature changes to 70 ∘C ,
bixtya [17]

Answer:

1528.3L

Explanation:

To solve this problem we should know this formula:

V₁ / T₁ = V₂ / T₂

We must convert the values of T° to Absolute T° (T° in K)

21°C + 273 = 294K

70°C + 273 = 343K

Now we can replace the data

1310L / 294K = V₂ / 343K

V₂ = (1310L / 294K) . 343K → 1528.3L

If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too

7 0
3 years ago
In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0
3 years ago
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