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3241004551 [841]
3 years ago
5

Newtons second law of motion states that

Chemistry
1 answer:
insens350 [35]3 years ago
3 0

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force,

Explanation:

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What is the electron configuration of the neutral lithium atom?
mote1985 [20]
<span>the electron configuration of the neutral Atom 
1s2 2s1
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8 0
4 years ago
When water is heated in an ocean the liquid water changes from and turns into?
sasho [114]
Hello!

The answer is Evaporation

When you add heat to water, it boils, releasing gas into the air. This is called Evaporation. The water changes from a liquid to a gas.

Hope this helped!
6 0
3 years ago
The work function is the energy that must be supplied to cause the release of an electron from a photoelectric material. The cor
vladimir2022 [97]

Answer:

No photoelectric effect is observed for Mercury.

Explanation:

From E= hf

h= Plank's constant

f= frequency of incident light

Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23

f= 11×10^14 Hz

The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface

3 0
4 years ago
The halpy of vaporization of H2O at 1 atm and 100 C is 2259 kJ/kg. The heat capacity of liquid water is 4.19 kJ/kg.C, and the he
Naddik [55]

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

        C_{p}_{liquid} = 4.19 kJ/kg ^{o}C

        C_{p}_{vaporization} = 1.9 kJ/kg ^{o}C

Heat of vaporization (\DeltaH^{o}_{vap}) at 1 atm and 100^{o}C is 2259 kJ/kg

        H^{o}_{liquid} = 0

Therefore, calculate the enthalpy of water vapor at 1 atm and 100^{o}C as follows.

            H^{o}_{vap} = H^{o}_{liquid} + \DeltaH^{o}_{vap}        

                                   = 0 + 2259 kJ/kg

                                   = 2259 kJ/kg

As the desired temperature is given 179.9^{o}C and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and 179.9^{o}C is calculated as follows.

             H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})

                             = 0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)

                              = 334.781 kJ/kg

Hence, enthalpy of water vapor at 10 bar and 179.9^{o}C is calculated as follows.

               H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})

             H^{D}_{vap} = 2259 kJ/kg + 1.9 \times (179.9 - 100)            

                              = 2410.81 kJ/kg

Therefore, calculate the latent heat of vaporization at 10 bar and 179.9^{o}C as follows.

       \Delta H^{D}_{vap} = H^{D}_{vap} - H^{D}_{liq}              

                         = 2410.81 kJ/kg - 334.781 kJ/kg

                         = 2076.029 kJ/kg

or,                      = 2076 kJ/kg

Thus, we can conclude that at 10 bar and 179.9^{o}C latent heat of vaporization is 2076 kJ/kg.

3 0
4 years ago
How many moles are in 25g of NaCI?
emmainna [20.7K]

Nacl = (. 23+35. 5)

= 58.5g

.

1 mol of Nacl = 58.5g

X mol Of Nacl = 25g

X Mol of Na Cl =25 ÷ 58.5

X mol of Nacl = 0.4 mol

4 0
3 years ago
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