Please , sorry if I'm asking too many questions

Please help with this question!

fenix001 [56]

H :) (I tried sorry if im wrong)

6 0

2 years ago

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What happens at night?

Alex

Warm air other the ocean moves onshore I think

7 0

3 years ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago

The anion O3− does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Indicate the va

Tju [1.3M]

One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.

Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.

We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.

Learn more: brainly.com/question/8646601

3 0

3 years ago

What number do you never write as a coefficient?

N76 [4]

Answer:

When no coefficient is written in front of a formula it is assumed to be 1. 4. Numbers appearing in the formulas are known as subscripts. These can never be changed when balancing the equation or you will change the identity of the substance.

Explanation:

1. Reactants go on the left hand side and products go on the right hand side of a chemical equation. Be sure to write the correct formulas for the reactants and products.

· Remember atoms are conserved in a chemical reaction. ie they are neither created or destroyed. So all the atoms in the reactants must end up somewhere among the products.

2. Count the number of atoms of each element, compound or ion in the reactants and products. If they are not equal proceed further.

3. Balance the atoms one at a time by placing coefficients in front of the formula so that the numbers of atoms of each element are equal on both sides of the equation. Remember atoms may exist in an element, compound or ion.

· It is usually easier to start with the atoms that occur in only one substance on each side of the equation.

· Balance the atoms that occur in compounds before attempting to balance atoms that occur in elemental form. e.g. H2, O2 or Cl2

· To make it easier if a polyatomic ion appears unchanged on both sides of the equation treat it as a whole unit.

· When no coefficient is written in front of a formula it is assumed to be 1.

4. Numbers appearing in the formulas are known as subscripts. These can never be changed when balancing the equation or you will change the identity of the substance.

· Remember with subscripts, any number to the right of parentheses multiplies each subscript within the parentheses.

eg Fe2(SO4)3 contains 2 Fe atoms, 3 S atoms and 12 O atoms.

5. Finally make sure that all the coefficients are in the smallest possible whole number ratio.

4 0

2 years ago