when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm
<h3>Conversion scale</h3>
14.6959 lb/in² = 1 atm
<h3>Data obtained from the question</h3>
- Pressure (in lb/in²) = 32.5 lb/in²
- Pressure (in ATM) =?
<h3>How to convert 32.5 lb/in² to atm</h3>
14.6959 lb/in² = 1 atm
Therefore
32.5 lb/in² = 32.5 / 14.6959
32.5 lb/in² = 2.21 atm
Thus, 32.5 lb/in² is equivalent to 2.21 atm
Learn more about conversion:
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An acid is deemed strong if it can readily or easy "donate" a proton (H+) to the other ions in the solutions. Also, to donate or lose the proton or H+, the acid must dissociate (split into ions) in the solution. The more it can readily dissociate, the stronger the acid is.
Answer: In the first paragraph, name a theme of Paul Laurence Dunbar's poem "Sympathy," and explain how it develops, citing specific examples
Explanation:
Answer:
Mass = 2.89 g
Explanation:
Given data:
Mass of NH₄Cl = 8.939 g
Mass of Ca(OH)₂ = 7.48 g
Mass of ammonia produced = ?
Solution:
2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O
Number of moles of NH₄Cl:
Number of moles = mass/molar mass
Number of moles = 8.939 g / 53.5 g/mol
Number of moles = 0.17 mol
Number of moles of Ca(OH)₂ :
Number of moles = mass/molar mass
Number of moles = 7.48 g / 74.1 g/mol
Number of moles = 0.10 mol
Now we will compare the moles of ammonia with both reactant.
NH₄Cl : NH₃
2 : 2
0.17 : 0.17
Ca(OH)₂ : NH₃
1 : 2
0.10 : 2/1×0.10 = 0.2 mol
Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.
Mass of ammonia:
Mass = number of moles × molar mass
Mass = 0.17 mol × 17 g/mol
Mass = 2.89 g
Answer:
Explanation:
An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.
In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.
The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.
