The by-product of the chlorination of an alkane is <u>HCl</u>
Explanation:
- Chlorination is the process of adding chlorine to drinking water to disinfect it and kill germs. Different processes can be used to achieve safe levels of chlorine in drinking water.
- Chlorination of alkane gives a mixture of different products.
- When consider mechanism of alkanes chlorination, free radicals are formed during the reaction to keep the continuous reaction.
- Different alkyl chloride compounds, extended carbon chains compounds and HCl are formed as products in product mixture.
- Chlorination byproducts, their toxicodynamics and removal from drinking water.
- Halogenated trihalomethanes (THMs) and haloacetic acids (HAAs) are two major classes of disinfection byproducts (DBPs) commonly found in waters disinfected with chlorine
- Chlorine is available as compressed elemental gas, sodium hypochlorite solution (NaOCl) or solid calcium hypochlorite (Ca(OCl)2
Answer:
119 kCal per serving.
Explanation:
The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:
Q = mcΔT
Q: heat energy
m: mass in g
c: specific heat capacity in cal/g°C
ΔT = temperature variation in °C
m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights 100g. Therefore:
Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C
Q = 1450 cal
1450 cal ____ 0.341 g peanuts
x ____ 28 g peanuts
x = 119061.58 cal
This means that the cal from fat per serving of peanuts is at least 119 kCal.
Answer:
One mole of carbon would look like 25/12.01
Explanation:
Firstly, you will divide 25 by 12.01 and get 2.081598
We know 1 mole equals the gram per atomic mass, so one mole of carbon is 12.01 grams. In conclusion, it would look like 25/12.01.
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
