Answer: The resultant pressure is 3.22 atm
Explanation:
Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.
(At constant volume and number of moles)
where,
= initial pressure of gas = 2.79 atm
= final pressure of gas = ?
= initial temperature of gas = 273K
= final temperature of gas = 315 K

Thus the resultant pressure is 3.22 atm
Answer:

Explanation:
Hello there!
In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:

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Answer:
a. Heat is often absorbed during reactions.
c. Heat is often released during chemical reactions.
Explanation:
Thermochemistry -
The study of thermochemistry involve the change in the amount of heat , during any physical or chemical process , is referred to as thermodynamics .
The focus of thermochemistry is on the changing amount of energy in the form of heat , on the system with respect to the surroundings .
The process like boiling , melting , sublimation , may require energy or releases energy , and hence are studied under thermochemistry .
Hence , from the given question ,
The correct options are - a , c.
Answer:
Total partial pressure, Pt = 821 mm Hg
Partial pressure of Helium, P1 = 105 mm Hg
Partial pressure of Nitrogen, P2 = 312 mm Hg
Partial pressure of Oxygen, P3 = ? mm Hg
According to Dalton's law of Partial pressures,
Pt = P1 + P2 + P3
So, <u>P3 = 404 mm Hg</u>
Answer:
3.6
Explanation:
Step 1: Given data
- Concentration of formic acid: 0.03 M
- Concentration of formate ion: 0.02 M
- Acid dissociation constant (Ka): 1.8 × 10⁻⁴
Step 2: Calculate the pH
We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.
![pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%3D%20-log%201.8%20%5Ctimes%2010%5E%7B-4%7D%20%2B%20log%20%5Cfrac%7B0.02%7D%7B0.03%7D%20%3D%203.6)