Answer:
Amount of Listerine can be prepared = 16.73 (Approx)
Explanation:
Given:
Listerine formula contains = 26.9% alcohol
Amount of alcohol = 4.5 L
Find:
Amount of Listerine can be prepared
Computation:
Amount of Listerine can be prepared = (Amount of alcohol) / (Listerine formula contains)
Amount of Listerine can be prepared = 4.5 / 26.9%
Amount of Listerine can be prepared = 16.73 (Approx)
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Answer:
The volume of hydrogen gas generated by given reaction is 9.48 L
Explanation:
Firstly, we calculate the mole of Zinc added, n = 25.5 / 65 = 0.392 mol
With the same equivalent, the mole of generated hydrogen gas will be the same, at 0.392 mol.
Secondly, we get the equation
<em>P . V = n . R . T</em>
with P (pressure, atm), V (volume, litre), n (mole, mol), R (ideal gas constant, 0.082), T (temperature, Kelvin)
So we have to convert the given information to correct unit
P = 742 mmHg = 0.976 atm
T = 15 °C = 288 K
Hence we can calculate the generated volume by
V = n . R . T / P = 0.392 x 0.082 x 288 / 0.976 = 9.48 L
CH3OCH3 and C2H5OH are isomers. They have same molecular formula. But with another formation.
C) the heat transfer to the boiler
