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otez555 [7]
3 years ago
5

when salt is added to water how does it affect the water's freezing points? Explain the science behind how this works at the par

ticle level for water and salt?
Chemistry
1 answer:
Sphinxa [80]3 years ago
3 0
Under normal conditions, water freezes at 0 degrees Celsius or 32 degrees Fahrenheit. However by, adding salt, the water will cause the freezing point to become lower.
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0.350 mol of a solid was dissolved in 260 mL of water at 21.2 oC. After the solid had fully dissolved, the final temperature of
Fittoniya [83]

Answer: Heat of the solution  = mass water × specific heat water × change in temperature

mass water = 260ml (1.00g/ml ) = 260g

specific heat of water = c(water) = 4.184J/ g°C

Heat change of water = final temperature - initial temperature

                                       = 26.5 - 21.2

                                        = 5.3 °C

H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J

Molar heat = \frac{5765J}{0.350mol}

                    = 16473J/mol

Explanation: finding molar heat requires first to look at  specific heat of water and the change of water temperature

7 0
3 years ago
Find the ratio of the effusion rate of hydrogen gas to the effusion rate of krypton gas.
Zielflug [23.3K]

This problem could be solved through the Graham’s law of effusion (also known as law of diffusion). This law states that the ratio of the effusion rate of the first gas and effusion rate of the second gas is equivalent to the square root of the ratio of its molar mass. Thus the answer would be 0.1098. 

6 0
2 years ago
Read 2 more answers
How did chemists make their proposed model of the sub atomic particles?
aalyn [17]

Answer:

Because they want to clarify or give solution to the particle that the other didn't know

Explanation:

Correct me if im wrong thanks

6 0
2 years ago
Determine the energy of 1.70 mol of photons for each of the following kinds of light. (Assume three significant figures.)PART A
BabaBlast [244]

<u>Answer:</u>

<u>For A:</u> The energy of the given amount of photons for infrared radiation is 1.271\times 10^5J

<u>For B:</u> The energy of the given amount of photons for infrared radiation is 4.026\times 10^5J

<u>For C:</u> The energy of the given amount of photons for infrared radiation is 1.355\times 10^6J

<u>Explanation:</u>

The relationship between energy and frequency is given by Planck's equation, which is:

E=n\rimes N_A\times \frac{hc}{\lambda}        ......(1)

where,

h = Planck's constant  = 6.62\times 10^{-34}Js

E = energy of the light

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

\lambda = wavelength of light

N_A = Avogadro's number = 6.022\times 10^{23}

n = number of moles of photons = 1.70 moles

Conversion factor used:  1m=10^9nm

  • <u>For A:</u>

Wavelength of infrared radiation = 1600nm=1.6\times 10^6m

Putting values in equation 1, we get:

E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-6}}\\\\E=1.271\times 10^5J

Hence, the energy of the given amount of photons for infrared radiation is 1.271\times 10^5J

  • <u>For B:</u>

Wavelength of visible light = 505nm=5.05\times 10^7m

Putting values in equation 1, we get:

E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{5.05\times 10^{-7}}\\\\E=4.026\times 10^5J

Hence, the energy of the given amount of photons for infrared radiation is 4.026\times 10^5J

  • <u>For C:</u>

Wavelength of ultraviolet radiation = 150nm=1.5\times 10^7m

Putting values in equation 1, we get:

E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.5\times 10^{-7}}\\\\E=1.355\times 10^6J

Hence, the energy of the given amount of photons for infrared radiation is 1.355\times 10^6J

3 0
3 years ago
cuanto electrones tiene que volver a perder cada elemento para lograr una configuración electrónica de gas noble?​
kozerog [31]

How many electrons does each element have to lose again to achieve a noble gas electron configuration?

Answer:

Explanation:

Each element will gain or lose electron to attain the octet of the noble gases in their outer shell electrons.

  • Most metals in group 1 and 2 will lose 1 and 2 electrons apiece to attain a noble configuration.
  • Non-metals are typically electronegative and will gain considerable amount of electrons to complete their octet.
  • Halogens will need one electron to complete their own out shell configuration.

Elements will gain or lose an amount of electron that will make it resemble noble gases.

5 0
2 years ago
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