Velocity = (distance covered in a direction) / (time to cover the distance)
Since the y-axis is position (distance), the speed at 3 seconds is the slope of the graph at 3 seconds, and NOT its y-value.
The slope is fairly easy to pick off, because the graph is so straight from 2 sec to 5 sec. During that time, the distance shrinks by 10 meters (from 10m to zero). So the slope of that whole piece of the graph is (-10m) / (3 sec).
That's a slope of (10/3 m/s) or 3.33 m/s .
In answer to the question, we can only give the speed at 3 sec, not the velocity, since we have no information about the direction of the motion. Consequently, I would call the speed a positive number. But it's not worth arguing about, so you should just select <em>choice-A </em>and not make a big scene.
As this mechanical energy is associated with height, it would be "Potential Energy" in particular.
U = mgh
U = F.h
U = 1.5 * 4
U = 6 Joules
So, 6 J of energy is lost before it hits the ground.
Hope this helps!
Explanation:
1) N₂ + O₂ → 2 NO
Kc = [NO]² / ([N₂] [O₂])
Set up an ICE table:
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CN_%7B2%7D%260.114%26-x%260.114-x%5C%5CO_%7B2%7D%260.114%26-x%260.114-x%5C%5CNO%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
Plug into the equilibrium equation and solve for x.
1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))
1.00×10⁻⁵ = (2x)² / (0.114 − x)²
√(1.00×10⁻⁵) = 2x / (0.114 − x)
0.00316 = 2x / (0.114 − x)
0.00361 − 0.00316x = 2x
0.00361 = 2.00316x
x = 0.00018
The volume is 1.00 L, so the concentrations at equilibrium are:
[N₂] = 0.114 − x = 0.11382
[O₂] = 0.114 − x = 0.11382
[NO] = 2x = 0.00036
2(a) Cl₂ → 2 Cl
Kc = [Cl]² / [Cl₂]
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CCl_%7B2%7D%262.0%26-x%262.0-x%5C%5CCl%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
1.2×10⁻⁷ = (2x)² / (2 − x)
1.2×10⁻⁷ (2 − x) = 4x²
2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²
2.4×10⁻⁷ ≈ 4x²
x² ≈ 6×10⁻⁸
x ≈ 0.000245
2x ≈ 0.00049
2(b) F₂ → 2 F
Kc = [F]² / [F₂]
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CF_%7B2%7D%262.0%26-x%262.0-x%5C%5CF%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
1.2×10⁻⁴ = (2x)² / (2 − x)
1.2×10⁻⁴ (2 − x) = 4x²
2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²
2.4×10⁻⁴ ≈ 4x²
x² ≈ 6×10⁻⁵
x ≈ 0.00775
2x ≈ 0.0155
F₂ dissociates more, so Cl₂ is more stable at 1000 K.
Question:
A) C6H6
B) CH3CH2CH2CH2CH2COH6
C) NaCl
D) NH3
Answer:
The correct option is;
A) C₆H₆
Explanation:
Heat of fusion = 6.02 kJ/mol
Heat of vaporization =40.8 kJ/mol
Here, we analyze each of the options as follows
A) C₆H₆
Benzene has a melting point of 5.5° C and a boiling point of
80.1 ° C similar to water
Heat of fusion = 9.92 kJ/mol
Heat of vaporization =30.8kJ/mol
B) CH₃CH₂CH₂CH₂CH₂COH₆
The above compound is more likely solid
C) NaCl solid
D) NH₃ melting point = -77.73 °C boiling point = -33.34 °C
Of the above, the compounds the one that closely resembles water is C₆H₆
DibujaZos da buenos tips e incluso el pinta con colores faber castel son económicos y el dibujo queda epico
