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mezya [45]
3 years ago
10

calculate the heat required in joules to convert 18.0 grams of water ice at a temperature of -20 c to liquid water at the normal

boiling point of water given specific heat of ice 2.09
Chemistry
1 answer:
KIM [24]3 years ago
7 0

Answer:

Explanation:

Heat required to convert ice to ice at 0⁰C

= mass x specific heat x rise in temperature

= 18 x 2.09 x 20

= 752.4 J .

heat required to convert ice at 0⁰C to water at 0⁰C

mass x latent heat of fusion

= 18 x 336

= 6048 J

Heat required to increase the temperature of water to 100⁰C

= 18 x 4.2 x 100

= 7560 J

Total heat required

7560 + 6048 + 752.4

= 14360.4 J

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What is atom.
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Answer:

atom is the smallest indivisible particle of an element

Explanation:

The symbol of 20 element are H He li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca

element is the simplest pure form of a substance .

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2 years ago
1.Vertical columns in the periodic table indicate what information? A.periods B.liquids C.synthetic elements D.groups and famili
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3 years ago
How much energy would be released as an electron moved from the n=4 to the n=3 energy level?
Thepotemich [5.8K]

The energy released when electron move from n=4 to n=3 is 0.66 eV

We know that in an atom energy of nth state is

                                     E_n =  -13.6/n^{2}  eV

where n is the energy level

Therefore,

                   E_4 = -13.6/4^{2} \\E_3  = -13.6/3^{2}

Thus, E_4  =   -0.85eV

         E_3  =  -1.51eV

Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

                                       E_4 -E_3

                                   =  -0.85 - ( -1.51)

                                   = 0.66eV

To know more about energy released in electron transition

brainly.com/question/8384785

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8 0
2 years ago
A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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