1 year ago

5. पहले अठारह तत्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के नियम कोलिखिए।

Chemistry

1 answer:

aniked [119]1 year ago

8 0

分門夌亥峉帕丑岶Explanation:

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What state of matter has the most kinetic energy

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Answer: Gas

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pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha

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Complete Question

Questions Diagram is attached below

Answer:

* $W=1142.86Joule$

* $Q=997.7J$

* $H=2140.5J$

Explanation:

From the question we are told that:

Temperature $T=337K$

Pressure $P=(60-55)Pa*10^5$

Volume $V=(1.6-1.4)m^3*10^{-3}$

Generally the equation for gas Constant is mathematically given by

$\frac{P_2}{P_1}=\frac{V_1}{V_2}^n$

$\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n$

$n=0.65$

Therefore

Work-done

$W=\int{pdv}$

$W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}$

$W=1142.86Joule$

Generally the equation for internal energy is mathematically given by

$Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}$

$Q=997.7J$

Therefore

$H=Q+W$

$H=997.7J-11.42.9$

$H=2140.5J$

5 0

2 years ago

A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d

Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

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2 years ago

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Georgia [21]

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2 years ago

Read 2 more answers

5. पहले अठारह तत्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के नियम कोलिखिए।​

5. पहले अठारह तत्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के नियम कोलिखिए।