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DIA [1.3K]
3 years ago
7

What does L represent? When did it form?

Chemistry
1 answer:
frosja888 [35]3 years ago
4 0

Answer:

All I can say is that it seems to go through the other layers. I would say this is an intrusive rock and probably the youngest

Explanation:

I suggest d i k e s. Put it together. somehow I cant post it when put together

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How much heat is absorbed during production of 112 g of NO by the combination of nitrogen and oxygen?
djverab [1.8K]

From the given chemical equation we see that 43 kcal of energy is needed for every 2 moles of NO. First let us calculate the moles of NO with a molar mass of 30 g/mol.

moles NO = 112 g / (30 g/mol) = 3.73 mol

 

So the total heat absorbed is:

heat = (43 kcal / 2 mol) * 3.73 mol

<span>heat = 80.195 kcal</span>

7 0
3 years ago
Explain why Earth and other planets were not solid when they formed during the beginning of the Precambrian, approximately 4,600
Crazy boy [7]
The beginning of the Pre Cambrian is known as the Hadean eon. The planet was just born, and it was very very hot, and in fact molten. There were a lot of volcanic activities, and because of the way the solar system was created, there were still a lot of collisions between different bodies
3 0
4 years ago
A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th
Bess [88]

Solution :

a). Applying the energy balance,

$\Delta E_{sys}=E_{in}-E_{out}$

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

           $ = 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

           = -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

           $ = 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

           = 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water} $

        = -9.09 + 10.76

         = 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

            = 63.94 -13.22

            = 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

            = 216.39 - 9677.95

            = -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} = $

$X_{destroyed} = X_{initial} - X_{final}$

                = 50.72 + 9461.56

                = 9512.22 kJ

6 0
3 years ago
Select the correct location on the image.
Mice21 [21]

Answer:

The third one from the left–the graduated cylinder.

Explanation:

The laboratory apparatus that gives an <em>"accurate" </em>or <em>"precise" </em>measurement of a liquid's volume is the<em> graduated cylinder</em>. All you have to do is to pour the liquid into the cylinder and read its measurement using the calibrated scale.

The graduated cylinder comes in different sizes, which means the scale divisions will depend on its size. When reading the measurement, it is important to take note to read at the <u><em>bottom of the meniscus</em></u> because it gives the most accurate volume.

7 0
3 years ago
Does all Newton's laws work together
9966 [12]
I believe so ya.....
6 0
3 years ago
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