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3 years ago

1325 mg to kg convert

Chemistry

1 answer:

Elis [28]3 years ago

3 0

Answer: 0.001325

Explanation: Hope that this helps :)

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I am confused, can somebody help me?

Ronch [10]

Answer:

electrons 79

neutrons 188

6 0

3 years ago

Read 2 more answers

Show that the Joule-Thompson Coefficient is zero for ideal gas.

melomori [17]

Answer:

Joule-Thomson coefficient for an ideal gas:

$\mu_{J.T} = 0$

Explanation:

Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.

Thus,

$\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H$

Also,

$H= H (T,P)$

$Differentiating\ it,$

$dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Also, $C_p$ is defined as:

$C_p = \left [\frac{\partial H}{\partial T}\right ]_P$

$So,$

$dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Acoording to defination, the ethalpy is constant which means $dH = 0$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H$

$Also,$

$\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p$

For an ideal gas,

$\left [\frac{\partial H}{\partial P}\right ]_T = 0$

So,

$0 =-\mu_{J.T}\times C_p$

Thus, $C_p$ ≠0. So,

$\mu_{J.T} = 0$

6 0

3 years ago

If the radius of copper is 0.128 nm, what is the pd of the (100) plane for copper in m-2?

kolbaska11 [484]

Copper has a FCC i.e. face centered cubic crystal structure. The 100 plane is essentially a planar section of the cubic cell where 4 Cu atoms occupy the 4 corners of the plane along with 1 Cu atom at the center of that plane. Each of the Cu atoms in the corners is shared by 4 adjacent unit cells. Thus, there are 2 Cu atoms present in the 100 plane (4*1/4 + 1 = 2).

Now, the planar density PD along the 100 plane is given as:

PD(100) = # atoms in the 100 plane/Area of 100 plane

= $\frac{2}{8R^{2} }$

Here R = radius = 0.128 nm = $0.128*10^{-9} m$

PD = $\frac{2}{8*(0.128*10^{-9}) ^{2} } = 1.526*10^{19} m^{-2}$

6 0

3 years ago

An ideal gas is kept in a 10-liter [L] container at a pressure of 2.5 atmospheres [atm] and a temperature of 310 kelvin [K].

Hatshy [7]

Answer:

5 L.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 10 L

Initial pressure (P1) = 2.5 atm

Final pressure (P2) = 5 atm

Final volume (V2) =.?

Since the temperature is constant, we shall apply the Boyle's law equation to determine the new volume of the gas. This can be obtained as follow:

P1V1 = P2V2

2.5 × 10 = 5 × V2

25 = 5 × V2

Divide both side by 5

V2 =25/5

V2 = 5 L

Thus, the new volume of the gas is 5 L

8 0

3 years ago

Can someone pls help me!!

Black_prince [1.1K]

Answer:

the answer is b

Explanation:

so the arrows signify the movement/ flow of energy from one organism to the next. so as one animal consumes another the energy from the the consumed animal flows through the consumer. its like eating a protein bar and the energy the protein bar gives you is what you have gotten from consuming the bar.

ps. i hope this helps

7 0

3 years ago