Answer:
The essence including its particular subject is outlined in the following portion mostly on clarification.
Explanation:
- The energy throughout the campfire comes from either the wood's latent chemical energy until it has been burned to steam up and launch up across the campfire. The electricity generation for something like a campfire seems to be in the context including its potential chemical energy which is contained throughout the firewood used only to inflame the situation.
- The energy output seems to be in the different types of heat energy radiating across the campfire, laser light generated off by the blaze, and perhaps a little number of electrical waves, registered throughout the firewood cracking whilst they combust throughout the blaze.
and,
chemical energy ⇒ heat energy + light energy + sound energy
<span>Electrons in a nitrogen-phosphorus covalent bond are not shared equally because nitrogen and phosphorus do not have the same electronegativity. The atoms spend more time around the most electronegative atom nitrogen.</span>
<u>Answer:</u> The new concentration of lemonade is 3.90 M
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of lemonade solution = 2.66 M
Volume of solution = 473 mL
Putting values in equation 1, we get:
Now, calculating the new concentration of lemonade by using equation 1:
Moles of lemonade = 1.26 moles
Volume of solution = (473 - 150) mL = 323 mL
Putting values in equation 1, we get:
Hence, the new concentration of lemonade is 3.90 M
Answer:
Abundance of 32S is 94.41%
Explanation:
The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:
Average atomic mass = ∑ Atomic mass istope*Abundance
For the sulfur:
32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>
<em>Where X is abundance of 32S and Y abundance of 33S</em>
Also we can write:
1 = X + Y + 0.0422 <em>(2)</em>
0.9578 - X = Y
Because the sum of the abundances = 1
Replacing (2) in (1):
32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422
32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344
-0.9435 = -0.99939X
0.9441 =X
In percentage, abundance of 32S is 94.41%
Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)
Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
