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3 years ago

Describe how substance move from one state to another

Physics

2 answers:

Ivan3 years ago

8 0

In large quantities, they're usually carried by the interstate trucking industry.

Mekhanik [1.2K]3 years ago

4 0

The temperature rises until the water reaches the next change of state — boiling. As the particles move faster and faster, they begin to break the attractive forces between each otherand move freely as steam — a gas. The process by which a substance moves from the liquid state to the gaseousstate is called boiling.

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A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius

taurus [48]

Torque acting dowward = 6 x 0.5 = 3 Nm

Torque acting to the right = 5 x 1 = 5 Nm

5 - 3 = 2 Nm

inertia = 1/2 mr^2

0.5 x 10 x 1^2 = 5 kg-m^2
2/5 = alpha = 0.4 rad /s^2

Hope this helps

5 0

3 years ago

Read 2 more answers

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago

What type of cloud is made up of both liquid water droplets & ice crystals

andre [41]

Answer: Altostratus clouds they are gray or blue gray mid level clouds.

Explanation:

7 0

1 year ago

How does mass affect the change in velocity of an object? Immersive Reader

Bas_tet [7]

Answer:

I would say the 3rd one

Explanation:

5 0

2 years ago

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average r

Dvinal [7]

Answer:

The lifetime of the particle is $\Delta t = 2.6*10^{-25} \ s$

Explanation:

From the question we are told that

The average rest energy is $E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J$

The intrinsic width is $\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J$

The lifetime is mathematically represented as

$\Delta t = \frac{h}{\Delta E}$

Where h is the Planck's constant with a value of $1.055*10^{-34} \ J\cdot s$

substituting values

$\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}$

$\Delta t = 2.6*10^{-25} \ s$

6 0

2 years ago