There are electrons between the carbon and oxygen atoms.
The electrons have a negative charge. The nuclei of the atoms are positively charged.
The electrical attraction between oppositely charged particles keeps the electrons between the atoms.
Answer:
8 × 10³ mol N₂
Explanation:
Step 1: Write the balanced equation
2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 5 H₂O + 0.5 O₂ + 3 N₂
Step 2: Convert the mass of C₃H₅N₃O₉ from tons to grams
We will use the conversion factor 1 t = 907185 g.
1.320 t × 907185 g/1 t = 1.197 × 10⁶ g
Step 3: Calculate the moles corresponding to 1.197 × 10⁶ g of C₃H₅N₃O₉
The molar mass of C₃H₅N₃O₉ is 227 g/mol.
1.197 × 10⁶ g × 1 mol/227 g = 5.27 × 10³ mol
Step 4: Calculate the moles of N₂ formed from 5.27 × 10³ moles of C₃H₅N₃O₉
The molar ratio of C₃H₅N₃O₉ to N₂ is 2:3.
5.27 × 10³ mol C₃H₅N₃O₉ × 3 mol N₂/2 mol C₃H₅N₃O₉ = 7.91 × 10³ mol N₂ ≈ 8 × 10³ mol N₂
S1 to GS as the amount of energy required to remove the electron is greatest when it is closest to the nucleus. think about two magents. does it take more energy to spereate them when they are far apart or when they are very close together? it's the same idea with electrons. the more energy required to remove, the more energy will be emmitted.
Hope u do well :) ik this maybe hard but you’ll get through it , the answer is D !
Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
![\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3DRT_b%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29-3.958%2B1.555lnP_c%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
Where,
is the Heat of vaoprization (J/mol)
is the normal boiling point of the gas (K)
is the Critical temperature of the gas (K)
is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:
Applying the above equation to find heat of vaporization as:
![\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3D8.314%5Ctimes307.4%20%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29-3.958%2B1.555ln36.4%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,
<u>Option C is correct</u>
