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2 years ago

A is a discussion characterized by procedures of argumentation.

Chemistry

2 answers:

katovenus [111]2 years ago

7 0

Answer is: debate.

Debate involves formal discussion on a particular topic where opposing arguments are put forward to argue for opposing viewpoints.

Debate is a formal type of discussion, often with a moderator and an audience

It happens in public meetings or academic institutions.

Debate is also an argument about a particular subject, in which many people are involved.

galina1969 [7]2 years ago

4 0

Debate is the correct answer
NOT counterclaim, rebuttal or resolution

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Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago

science-what kind of input #o you need to provide to make the pen work. Describe the process and out put that result from the in

seropon [69]

Answer:

10 1/3

Explanation:I looked it UP

3 0

2 years ago

A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of

pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

pH = 10.05

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = 0.20

I hope it helps!

3 0

2 years ago

What is the net ionic equation for the reaction if any that occurs when aqueous solutions of Na2CO3 and HCL are mixed?

strojnjashka [21]

Answer:

CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)

Explanation:

The balanced reaction between Na2CO3 and HCl is given as;

Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)

The next step is o express the species as ions.

The complete ionic equation for the above reaction would be;

2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)

The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.

The spectator ions are; Na⁺ and Cl⁻

The net ionic equation is given as;

CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)

7 0

2 years ago

What is the transfer of energy by electro magnetic waves??

gladu [14]

It's called radiation

4 0

3 years ago

Read 2 more answers