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juin [17]
3 years ago
10

Hey conner i miss you soooooo much and does anyone have tik tok

Chemistry
2 answers:
irakobra [83]3 years ago
8 0

Answer:

i have tik tok

Explanation:

kherson [118]3 years ago
6 0

Answer:

I have tik tok

Explanation:

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The combining of two atomic nuclei to produce one larger nucleus is called
Naddika [18.5K]
C.
Nuclear fission involves the high energy collision of two atomic nuclei to produce one larger nucleus; this process powers stars.
6 0
4 years ago
3. What is the shape of 3p atomic orbital? (1 point)
sukhopar [10]

Option B

dumbbell is the shape of 3p atomic orbital

<u>Explanation:</u>

Atomic orbitals are three-dimensional places inside an atom where there is a tremendous chance of detecting electrons.  The p orbital, which develops in complexity and ought 2 spaces encompassing the atom core, or is defined as possessing a dumbbell pattern. The 3p atomic orbital is at energy level 3, as seen by the number 3 filed ere the character.

These orbitals have identical appearances but are arranged asymmetrically in location. p orbitals are wavefunctions with l=1. They ought an angular frequency that is ununiform at each angle. They have an appearance that is much defined as a "dumbbell".

4 0
3 years ago
Consider the equilibrium
vladimir1956 [14]

Answer:

Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}

\Delta _rG=1.01x10^5J/mol

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar

Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}

Finally, the Gibbs free energy for the reaction at 298.15K is:

\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol

Best regards.

3 0
4 years ago
Which answer best describes this reaction?
Maksim231197 [3]
The photos for the reaction arent here so ill insert them below 

3 0
3 years ago
Read 2 more answers
American cars use about 600,000,000 gallons of oil per year. How many liters of oil do American cars use per year? Report your a
kakasveta [241]

Answer:

2400000000

Explanation:

i believe that this should be the correct answer i punched in the numbers of conversion based off of what you put in your question.

4 0
3 years ago
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