Answer:
C3F7
'Tri' means 3
Tricarbon means 3 carbon
"Hepta" means 7
Fluoridez molecular formula is "F"
hence F7
Explanation:
The given data is as follows.
Mass of apple sauce mixture = 454 kg
Heat added (Q) = 121300 kJ
Heat capacity (
) of apple sauce at
= 4.0177
So, Heat given by heat exchanger = heat taken by apple sauce
Q = 
or, Q =
Putting the given values into the above formula as follows.
Q =
121300 kJ = 
= 
Thus, we can conclude that outlet temperature of the apple sauce is
.
The answer is: when the aim is to show electron distributions in shells
An orbital notation is more appropriate if you want to show how the electrons of an atom are distributed in each subshell. This is because there are some atoms that have special electronic configurations that aren't obvious in just written configurations.
The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
![pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247](https://tex.z-dn.net/?f=pKa%3D-log%5BH%5D%20%3D%20-%20log%20%5B%205.66%20%2A%2010%5E%7B-7%7D%5D%5C%5C%20%5C%5Cpka%20%3D%207%20-%20log%20%285.66%29%3D7-0.753%3D6.247%5C%5C%5C%5Cpka%20%3D%206.247)
The pH of the buffer can be known as
![pH = pK_{a} + log[\frac{[A-]}{[HA]}}]](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5B%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%7D%5D)
The concentration of ![[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20Moles%20of%20%5BA%5D%2FTotal%20volume%20%3D%200.608%2F2%20%3D%200.304%20M%5C%5C)
Similarly, the concentration of [HA] = 
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]

So, the pH of the buffer is 6.1236.
Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole