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2 years ago

What is the [h3o + ] in a 0.050 m solution of ba(oh)2?

2 answers:

7 0

The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below

write the equation for the dissociation of Ba(OH)2

Ba(OH)2 = Ba^2+ +2OH^-

calculate the OH- concentration

by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M

by use of the formula ( H3O+)(OH-) = 1 x10 ^-14

by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-

substitute for OH-

H3O+ = (1 x10^-14 )/0.1

= 1 x10^-3 M

Ede4ka [16]2 years ago

4 0

Answer is: concentration of hydronium ions are 10⁻¹³ M.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.050 M.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
c(OH⁻) = 0.10 M = 10⁻¹ M.
c(OH⁻) · c(H₃O⁺) = 1·10⁻¹⁴ M².
c(H₃O⁺) = 10⁻¹⁴ M² ÷ 10⁻¹ M.
c(H₃O⁺) = 10⁻¹³ M.

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The successive ionization energies of a certain element are I1 = 589.5 kJ/mol, I2 =1145 kJ/mol, I3= 4900 kJ/mol, I4 = 6500 kJ/mo

Gnom [1K]

Answer:

Explanation:

For any element to exhibit the pattern of ionization energy shown in the question, it must possess five electrons in its outermost shell. These five electrons are not lost at once. They are lost progressively until the valence shell becomes empty. The ionization energy increases steadily as more electrons are lost from the valence shell.

The only pentavalent element among the options in arsenic, hence the answer.

7 0

2 years ago

Read 2 more answers

A 60.2-ml sample of hg (density = 13.6 g/ml) contains how many atoms of hg?

Over [174]

Density is calculated using the following rule:
density = mass / volume
therefore:
mass = density * volume
mass of Hg = 13.6 * 60.2 = 818.72 grams

From the periodic table:
molar mass of Hg = 200.59 grams

number of moles = mass / molar mass
number of moles of Hg = 818.72 / 200.59 = 4.08 moles

each mole contains Avogadro's number of atoms.
Therefore,
number of atoms in the given sample = 4.08 * 6.022 * 10^23
= 2.456976 * 10^24 atoms

6 0

2 years ago

A 23.74-mL volume of 0.0981 M NaOH was used to titrate 25.0 mL of a weak monoprotic acid solution to the stoi- chiometric point.

Dmitrij [34]

Answer:

Molar concentration of the weak acid solution is 0.0932

Explanation:

Using the formula: $\frac{C_aV_a}{C_bV+b} = \frac{n_a}{n_b}$

Where Ca = molarity of acid

Cb = molarity of base = 0.0981 M

Va = volume of acid = 25.0 mL

Vb = volume of base = 23.74 mL

na = mole of acid

nb = mole of base

Since the acid is monopromatic, 1 mole of the acid will require 1 mole of NaOH. Hence, na = nb = 1

Therefore, $C_a = \frac{C_bV_b}{V_a}$

Ca = 0.0981 x 23.74/25.0

= 0.093155 M

To 4 significant figure = 0.0932 M

3 0

2 years ago

Read 2 more answers

En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello

MAXImum [283]

Answer:

a) $m_{CH_4}=2630kg$

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

$Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ$

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

$Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}$

$m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg$

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

$n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol$

Con ello, los kilogramos de metano que cuestan 0,45 €:

$44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4$

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0

2 years ago

How many moles of carbon are in 7.87x10^7 carbon molecules?

zaharov [31]

There are 6.022*10^23 molecules in 1 mole of carbon
So how many will moles will be 7.87*20^7?
Let the required number of moles be ‘x’.
1 mole ———6.022*10^23
x moles———7.87*10^7
(Cross multiplication)
x=7.87*10^7/6.022*10^23
Therefore x=1.3*10^-16

6 0

1 year ago