Structure 2 is the nucleus.
Answer:
Follows are the solution to this question:
Explanation:
Given value:
![v_1=1m^3 \\ t_1 = 600 \ K \\ p_1 = 1000 \ kpa \\v_1 = 5 \ v_1 = 5 \ m^3](https://tex.z-dn.net/?f=v_1%3D1m%5E3%20%5C%5C%20t_1%20%3D%20600%20%5C%20K%20%5C%5C%20p_1%20%3D%201000%20%5C%20kpa%20%5C%5Cv_1%20%3D%205%20%5C%20v_1%20%20%3D%205%20%5C%20m%5E3)
In point a:
Calculating the process of Isothermal, when the temperature is constant:
![\to T_1 = T_2 = 600 \ K \\\\\to \bold{P_1V_1 = P_2V_2} \\\\\to 1000 \ (Kpa) \times 1 \ (m^3) \neq p_2 \times 5 \ (m^3)\\\\\to p_2 = 200 \ kpa\\\\\to w = nRT \ In (\frac{v_2}{v_1}) = p_1v_1 \ ln ( \frac{v_2}{v_1})](https://tex.z-dn.net/?f=%5Cto%20%20T_1%20%3D%20T_2%20%3D%20600%20%5C%20K%20%5C%5C%5C%5C%5Cto%20%5Cbold%7BP_1V_1%20%3D%20P_2V_2%7D%20%5C%5C%5C%5C%5Cto%201000%20%5C%20%28Kpa%29%20%20%5Ctimes%201%20%5C%20%28m%5E3%29%20%5Cneq%20p_2%20%5Ctimes%20%205%20%5C%20%28m%5E3%29%5C%5C%5C%5C%5Cto%20p_2%20%3D%20200%20%5C%20kpa%5C%5C%5C%5C%5Cto%20w%20%3D%20nRT%20%5C%20In%20%28%5Cfrac%7Bv_2%7D%7Bv_1%7D%29%20%3D%20%20p_1v_1%20%20%20%5C%20ln%20%28%20%5Cfrac%7Bv_2%7D%7Bv_1%7D%29)
![= 1000 \times 1 \times ln (\frac{5}{1}) \\\\ = 1.61 \ KJ](https://tex.z-dn.net/?f=%3D%201000%20%5Ctimes%20%201%20%5Ctimes%20ln%20%20%28%5Cfrac%7B5%7D%7B1%7D%29%20%5C%5C%5C%5C%20%3D%201.61%20%5C%20KJ)
In point b:
Calculating the adiobatic process:
![\to p_1v_1^\gamma = p_2v_2^\gamma \\\\ \to \gamma = \frac{c_p}{c_v} \\\\\to R= c_p -c_v \\\\ \to c_p= 21 \frac{J}{mol.k}\\\\\to \gamma = \frac{c_p}{c_p-R} \\\\](https://tex.z-dn.net/?f=%5Cto%20%20p_1v_1%5E%5Cgamma%20%20%3D%20p_2v_2%5E%5Cgamma%20%5C%5C%5C%5C%20%5Cto%20%5Cgamma%20%20%3D%20%5Cfrac%7Bc_p%7D%7Bc_v%7D%20%5C%5C%5C%5C%5Cto%20%20R%3D%20c_p%20-c_v%20%20%5C%5C%5C%5C%20%5Cto%20c_p%3D%2021%20%5Cfrac%7BJ%7D%7Bmol.k%7D%5C%5C%5C%5C%5Cto%20%5Cgamma%20%20%3D%20%5Cfrac%7Bc_p%7D%7Bc_p-R%7D%20%5C%5C%5C%5C)
![= \frac{21}{21.8}\\\\ = 1.62\\\\= 1000 \times 1^{1.62}\\\\ = p_2 \times 5^{1.62}\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B21%7D%7B21.8%7D%5C%5C%5C%5C%20%3D%201.62%5C%5C%5C%5C%3D%201000%20%5Ctimes%201%5E%7B1.62%7D%5C%5C%5C%5C%20%3D%20p_2%20%5Ctimes%20%205%5E%7B1.62%7D%5C%5C%5C%5C)
![p_2 = 73.73 \ Kpa](https://tex.z-dn.net/?f=p_2%20%3D%20%2073.73%20%5C%20Kpa)
![\to p_1^{1-\gamma} t_1^{\gamma} = p_2^{1-\gamma } t_2^{\gamma }](https://tex.z-dn.net/?f=%5Cto%20p_1%5E%7B1-%5Cgamma%7D%20t_1%5E%7B%5Cgamma%7D%20%3D%20p_2%5E%7B1-%5Cgamma%20%7D%20t_2%5E%7B%5Cgamma%20%7D)
![= 1000^{1-1.62} \times 600^{1.62} = 73.73^{1-1.62} \times t_2^{\gamma}\\\\ \to t_2^{\gamma} = 6288.5\\\\\to t_2= 6258.5 ^\frac{1}{1.62} \\\\](https://tex.z-dn.net/?f=%3D%201000%5E%7B1-1.62%7D%20%5Ctimes%20600%5E%7B1.62%7D%20%3D%2073.73%5E%7B1-1.62%7D%20%5Ctimes%20t_2%5E%7B%5Cgamma%7D%5C%5C%5C%5C%20%5Cto%20t_2%5E%7B%5Cgamma%7D%20%3D%206288.5%5C%5C%5C%5C%5Cto%20t_2%3D%206258.5%20%5E%5Cfrac%7B1%7D%7B1.62%7D%20%5C%5C%5C%5C)
![= 221.2 \ k](https://tex.z-dn.net/?f=%3D%20221.2%20%5C%20k)
In point c:
![\to w= \frac{p_2v_2-p_1v_1}{\gamma -1}](https://tex.z-dn.net/?f=%5Cto%20w%3D%20%5Cfrac%7Bp_2v_2-p_1v_1%7D%7B%5Cgamma%20-1%7D)
![= \frac{(73.73 \times 5)- ( 1000\times 1)}{1.620-1}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%2873.73%20%5Ctimes%205%29-%20%28%201000%5Ctimes%201%29%7D%7B1.620-1%7D)
![= \frac{( -631.35 )}{.620}\\\\= -1018.31 \ KJ](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%20-631.35%20%29%7D%7B.620%7D%5C%5C%5C%5C%3D%20-1018.31%20%5C%20KJ)
workdone by gas is 1.018 KJ
A vestigial structure is a feature of an organism's body that was once an adaptation shaped by natural selection, but which is no longer useful in their current environment. For example, some species of fish that live in completely dark caves have eyes, although their eyes cannot see and serve no function.
KCO₂
K₂C₂O₄
Explanation:
Given parameters:
Percent composition:
K = 47%
C = 14.5%
O = 38.5%
Molar mass of compound = 166.22g/mol
Unknown:
Empirical formula of compound = ?
Molecular formula of compound = ?
Solution:
The empirical formula of a compound is its simplest formula. Here is how to solve for it:
K C O
Percent
composition 47 14.5 38.5
Molar mass 39 12 16
Number
of moles 47/39 14.5/12 38.5/16
moles 1.205 1.208 2.4
Dividing
by smallest 1.205/1.205 1.208/1.205 2.4/1.205
1 1 2
Empirical formula KCO₂
Molecular formula
This is the actual combination of the atoms:
Molecular formula = ( empirical formula of KCO₂)ₙ
Molar mass of empirical formula = 39 + 12 + 2(16) = 83g/mol
n factor = ![\frac{true molecular mass}{molar mas of empirical formula}](https://tex.z-dn.net/?f=%5Cfrac%7Btrue%20molecular%20mass%7D%7Bmolar%20mas%20of%20empirical%20formula%7D)
n factor =
= 2
Molecular formula of compound = ( KCO₂)₂ = K₂C₂O₄
Learn more:
Empirical formula brainly.com/question/2790794
#learnwithBrainly
Answer:chemical markers
Explanation:
I’m smart give brainlest plz