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
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The percent composition of this compound :
Mg = 72.182%
N = 27.818%
<h3>Further explanation</h3>
Given
9.03 g Mg
3.48 g N
Required
The percent composition
Solution
Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements
Total mass of the compound :
= 9.03 g + 3.48 g
= 12.51 g
The percent composition :
Mg : 9.03/ 12.51 g x 100% = 72.182%
N : 3.48 / 12.51 g x 100% = 27.818%
The copper wire was sanded before burning in order to make sure that copper metal was exposed on the surface of the wire.
Answer: B
Explanation
The copper wire when placed in atmosphere without coating leads to oxidation of copper metal with respect to the impurities present in the atmosphere.
As copper is electropositive in nature, so electronegative ions present in the universe will try to react with copper and the copper will react easily with other elements.
So generally copper wire is coated with color or polymer coating.
In this case, the copper wire without any coating is sanded, so that the eddy sheets or polishing materials on friction with copper wire will remove the impurities by the electrostatic law of conservation of charges and charge transfer.
As the impurities are removed when copper wire is sanded, the copper atoms will be exposed on the surface of the wire leading to burning of copper in the copper wire.
Cl2 is nonpolar so it has to be only London dispersion force (LDF)
Answer:
11%
Explanation:
1) Calculate van 't Hoff factor:
Δt = i Kf m
0.31 = i (1.86) (0.15)
i = 1.111
2) Calculate value for [H+]:
CCl3COOH ⇌ H+ + CCl3COO¯
total concentration of all ions in solution equals:
(1.11) (0.15) = 0.1665 m
This is a molality, but we will act as if it a molarity since we will assume the density of the solution is 1.00 g/cm3, which makes the molarity equal to the molality.
0.1665 = (0.15 − x) + x + x
x = 0.0165 M
3) Calculate the percent dissociation:
0.0165/ 0.15 = 11 %
