Answer;
C7H14O2
Solution;
Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)
Mass of carbon = 12/44 × 2.726 g
= 0.743455 g
Mass of Hydrogen = 2/18 × 1.116 g
= 0.124 g
Mass of oxygen = 1.152 - (0.7435 + 0.124)
= 0.2845 g
Moles of carbon ; 0.7435/12 = 0.06196 moles
Moles of hydrogen; 0.124/1 = 0.124 moles
Moles of oxygen; 0.2845/16 = 0.01778 moles
Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778
= 3.5 : 7.0 : 1
To make them whole numbers ; we multiply the ratios by 2 to get;
(3.5 : 7.0 : 1 )2 = 7 : 14 : 2
Thus, the empirical formula of Isobutyl propionate is C7H14O2
<u>Answer:</u> The mass of chlorine needed by the plant per day is
<u>Explanation:</u>
We are given:
Volume o water treated per day = 25,000,000 gallons
Converting this volume from gallons to liters, we use the conversion factor:
1 gallon = 3.785 L
So,
Amount of chlorine applied for disinfection = 10 mg/L
Applying unitary method:
For 1 L of water, the amount of chlorine applied is 10 mg
So, for of water, the amount of chlorine applied will be
Hence, the mass of chlorine needed by the plant per day is
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what are you doing and how is your day ?
Answer:
The answer to your question is: letter E
Explanation:
Normally, the correct order of boiling points is:
Alcohols > Ketones > Ether > Alkane
Then
A. n-butane < 1-butanol < diethyl ether < 2-butanone
B. n-butane < 2-butanone < diethyl ether < 1-butanol
C. 2-butanone < n-butane < diethyl ether < 1-butanol
D. n-butane < diethyl ether < 1-butanol < 2-butanone
E. n-butane < diethyl ether < 2-butanone < 1-butanol
(- 1°C) < 34.6°C < 79.64°C < 117.7°C