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4 years ago

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7. A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is its density

?

1 answer:

BaLLatris [955]4 years ago

5 0

Answer:

$d=1.6\ g/cm^3$

Explanation:

Given that,

The dimension of a square chunk of plastic is 5 cm × 5 cm × 5 cm

Mass, m = 200 g

We need to find its density

Density = mass/volume

It will form a cube. It volume = side³

So,

$d=\dfrac{200\ g}{5^3\ cm^3}\\\\d=1.6\ g/cm^3$

Hence density is $1.6\ g/cm^3$

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In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌

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Answer: Equilibrium constant is 0.70.

Explanation:

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Initial concentration of $H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M$

equilibrium concentration of $CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M$ [/tex]

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial conc. 0.35 M 0.40M 0 0

At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}$

$K_c=\frac{x\times x}{(0.40-x)(0.35-x)}$

we are given : (0.35-x)= 0.18

x = 0.17

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

$K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}$

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Answer:

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Explanation:

The device which is used to accelerate charged particles to higher energies is called a cyclotron. It is based on the principle that the particle when placed in a magnetic field will possess a magnetic force. Just because of this Lorentz force it moves in a circular path.

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The work done by the particles is equal to the kinetic energy stored in it such that,

$qV=\dfrac{1}{2}mv^2$

v is the speed with which the particles enter the cyclotron

So,

$v=\sqrt{\dfrac{2qV}{m}}$

So, the speed with which the particles enter the cyclotron is $v=\sqrt{\dfrac{2qV}{m}}$ . Hence, this is the required solution.

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If the concentration of h uons were to decrease what would happen to the oh ions

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4 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

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Therefore, mass of water = Density × Volume

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Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

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3 years ago