This should assist you with your inquiry: Gravitational acceleration = 9.8 m/s². The formula is 1/2 x 9.8 x 7² = 240.1 m, indicating that the rock has fallen a distance of 240.1 m. I can't assist with additional details, but this should provide a foundation.
Answer:
The partial pressure of SO₃ is measured at 82.0 atm.
Explanation:
The equilibrium constant Kp is defined as the ratio of the equilibrium pressures of the gaseous products, each raised to the power of their respective coefficients in the reaction, divided by the pressures of the gaseous reactants raised to their coefficients.
For the given reaction,
2 SO₂(g) + O₂(g) → 2 SO₃(g)
![Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm](https://tex.z-dn.net/?f=Kp%20%3D%200.345%20%3D%20%5Cfrac%7B%28pSO_%7B3%7D%29%5E%7B2%7D%20%7D%7B%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%28pSO_%7B2%7D%29%5E%7B2%7D%20%5Ctimes%20pO_%7B2%7D%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%20%5Csqrt%5B%5D%7B0.345%20%5Ctimes%20%2835.0%29%5E%7B2%7D%20%5Ctimes%2015.9%20%7D%20%5C%5CpSO_%7B3%7D%20%3D%2082.0%20atm)
Answer:
During this process, energy is released as a bond is formed.
Explanation:
The elements present on the left side of the reaction equation are known as reactants, while those on the right are called the products.
In this case, N(g) and N(g) are the reactants, and N2(g) along with energy are the products.
Post-reaction, N(g) has vanished and has been replaced by N2(g). It confirms that a bond exists between the two N(g).
Moreover, energy has been released as part of the products (indicated on the right side)
The accurate statement is ''A bond is formed and energy is released''
Answer: The molecular formula will be 
Explanation:
When percentages are provided, we assume the total mass to be 100 grams.
Thus, the mass of each element corresponds to the specified percentage.
Mass of C= 70.6 g
Mass of H = 5.9 g
Mass of O = 23.5 g
Step 1: convert given masses to moles.
Moles of C =
Moles of H =
Moles of O =
Step 2: For determining the mole ratio, divide each molar amount by the smallest number of moles calculated.
For C = 
For H = 
For O =
The resulting ratio of C: H: O= 4: 4: 1
Hence, the empirical formula obtained is 
The empirical weight is calculated as
= 4(12)+4(1)+1(16)= 68g.
The molecular weight = 136 g/mole
Now the molecular formula needs to be obtained.

The molecular formula can be derived as=