A second order reaction varies with the square of the concentration of the reactant. Therefore, halving the concentration will reduce the rate of reaction by a factor of 4.
The answer is E.
Answer: OH−.
Explanation: Hydroxide, any chemical compound containing one or more groups, each comprising one atom each of oxygen and hydrogen bonded together and functioning as the negatively charged ion OH-.
Answer:
Explanation:
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In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:
In terms of mass, specific heat and temperature change is:
Now, solve for the final temperature, as follows:
Then, plug in the masses, specific heat and temperatures to obtain:
Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.
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Protons always has a positive charge
Answer:
T° freezing solution → -11.3°C
T° boiling solution → 103.1 °C
Explanation:
Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"
This salt dissociates as this:
SnCl₄ (aq) → 1Sn⁴⁺ (aq) + 4Cl⁻ (aq) (so i =5)
The formula for the colligative property of freezing point depression and boiling point elevation are:
ΔT = Kf . m . i
where ΔT = T° freezing pure solvent - T° freezing solution
ΔT = Kb . m . i
where ΔT = T° boiling solution - T° boiling pure solvent
Freezing point depression:
0° - T° freezing solution = 1.86°C/m . 1.22 m . 5
T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C
Boiling point elevation:
T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5
T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C