Respuesta:
968 g Ca(OH)₂
Explicación:
Paso 1: Calcular la masa de solución
Tenemos 1500 mL de una solución cuya densidad es 1.17 g/mL, es decir, 1 mL de solución tiene una masa de 1.17 g.
1500 mL × 1.17 g/mL = 1.76 × 10³ g
Paso 2: Calcular la masa de hidróxido de calcio en 1.76 × 10³ g de solución
La solución tiene una concentración de 55% en masa de hidróxido de calcio, es decir, cada 100 gramos de solución hay 55 gramos de hidróxido de calcio.
1.76 × 10³ g Solución × 55 g Ca(OH)₂/100 g Solución = 968 g Ca(OH)₂
<u>Answer:</u> The molality of 
solution is 0.782 m
<u>Explanation:</u>
Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:
.....(1)
Given values:
Moles of = 0.395 mol
Mass of solvent (water) = 0.505 kg
Putting values in equation 1, we get:
Hence, the molality of solution is 0.782 m
Answer: The correct option is The properties of a noble gas.
Explanation: There are 7 periods in the periodic table.
The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).
- The electronic configuration for Helium is
. For He, The outermost electrons are 2.
- The electronic configuration for all the other elements is
( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.
All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.
Therefore, the last element of each period always have the properties of a noble gas.
There are three perfect squares in a standard die; 1, 2, 4. If there is two standard dies, then the probability of getting a perfect square is 1/3 x 1/3 = 1/9.
There are 4 numbers less than 5 in a standard die, making it 1/4 x 1/4=1/16.
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
