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Answer:
aquatic life will be harmed by ammonia
it causes significant losses in production
The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ
= i
m
where:
Δ
= change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration =
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ
= i
m we get,
Δ
= 2 × 1.86 × 0.1219
Δ
= 0.4536°C
So, Δ
of solution is,
Δ
= 0.00°C - 0.45°C
Δ
= - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Learn more about freezing point here;
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1) Chemical equation:
2Ca + O2 ---> 2CaO
2) molar ratios
2 mol Ca : 1 mol O2 : 2 mol CaO
3) Convert 35.4 g of Ca to moles:
number of moles = mass in grams / atomic mass = 35.4 g / 40.078 g/mol = 0.883 mol
3) use proportion and solve for x:
2 mol Ca / 1 mol O2 = 0.883 mol Ca / x
=> x = 0.883 mol Ca * 1 mol O2 / 2 mol Ca
=> x = 0.4416 mol O2
3) use ideal gases equation to transform moles to liters
pV = nRT => V = nRT / p =
V = [0.4416 mol * 0.0821 atm*liter / atm*K * 285K] / 1.2 atm = 8.6 liter
Answer: 8.6 liters
Answer: The empirical formula for C6H12O6 is CH2O. Every carbohydrate, be it simple or complex, has an empirical formula CH2O
Explanation: