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likoan [24]
2 years ago
6

Consider the titration of 30.0 mL of 0.200 M HClO4 with 0.100 M KOH. Calculate the pH of the resulting solution after the follow

ing volumes of KOH have been added.
0.0 mL pH is
15.0 mL pH is
30.0 mL pH is
60.0 mL pH is
70.0 mL pH is
Chemistry
1 answer:
Umnica [9.8K]2 years ago
8 0

Answer:

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The movement of fluid materials caused by density differences
Dovator [93]
This movement is known as convection or convection currents. This occurs due to the fact that warmer fluid is of lower density than colder fluid. This causes warmer fluid to rise and colder fluid to sink. This creates circulatory currents within the body of the fluid.
8 0
2 years ago
Find the molality of the solution if 42 grams of lithium chloride (LiCl) are dissolved in 3.6 kg of water.
omeli [17]
The molality is calculated using the following rule:
molality = number of moles of solute / kg of solvent

From the periodic table:
molar mass of lithium = 6.941 gm
molar mass of chlorine = 35.453 gm

molar mass of LiCl = 6.941 + 35.453 = 42.394 gm
number of moles found in 42 gm = mass / molar mass = 42 / 42.394 = 0.99

molality = 0.99 / 3.6 = 0.275 m
3 0
3 years ago
What is the volume of an irregular object if the initial volume in the graduated cylinder is 3.50 ml and it rises to 7.50 ml aft
Margarita [4]
Answer is: the volume of an irregular object is 4,00 ml.
<span>Volume is the amount of space the object occupies and can be finded immersing it in water in a container with volume markings and than see how much the level of the container changes (goes up). 
</span>V(irregular object) = V(final volume) - V(initial volume).
V(irregular object) = 7,50 ml - 3,50 ml.
V(irregular object) = 4,00 ml.
8 0
3 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
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