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2 years ago

Using a ph meter, you find the ph of an unknown solution to be 8. 0. How would you describe this solution?.

Chemistry

1 answer:

insens350 [35]2 years ago

4 0

Answer:

Just Barely Base/Neutral

Explanation:

a pH of 8.0 is greater than Neutral (7.0) but is still neutral due to it being more neutral than a base

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PLS HELP ASAP I NEED HELP IMAGE BELOW

andriy [413]

I think it would be 1, 4, 5! Also I love your profile picture!

5 0

2 years ago

Read 2 more answers

Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T

GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0

3 years ago

How to set up the rate expressions for the following mechanism?

Anestetic [448]

Answer:

Explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law = $\dfrac{d[D]}{dt}=k_3[B][C] --- (1)$

$\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]$

Using steady-state approximation;

$\dfrac{d[B]}{dt}=0$

$k_1[A]-k_2[B]-k_3[B][C] = 0$

$[B] = \dfrac{k_1[A]}{k_2+k_3[C]}$

From equation (1), we have:

$\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}$

when the pressure is high;

k₂ << k₃[C]

$\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \ \text{first order}$

k₂ >> k₃[C]

$\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \ \text{second order}$

3 0

3 years ago

What will be the end product for each electrode after electrolysis if the solution is concentrated aqueous sodium chloride ?

Eddi Din [679]

Answer:

Cathode: Hydrogen gas

Anode: Chlorine gas

Explanation:

① Write down the ions present in the electrolyte

Cations: Na⁺, H⁺

Anions: Cl⁻ , OH⁻

② Decide which ions are preferentially discharged.

These are the factors:

For the discharge of cations,

- Reactivity series

(The lower the position of the cation in the reactivity series, the easier it is to be discharged)

For the discharge of anions,

- Concentration effect (look at this first)

(The more concentrated the ion, the easier for it to be discharged)

- If solution is not concentrated (dilute), look at the position of the anion on the electrochemical series.

(The lower the position of the anion on the electrochemical series, the easier of it to be discharged)

In this case:

For cations, H⁺ ions are selectively discharged at the cathode as its position is lower than Na⁺ in the reactivity series.

Anion: Cl⁻ ions, being more concentrated, are selectively discharged at the anode.

☆For electrolysis,

Cation at the cathode (-ve terminal)

Anion at the anode (+ve terminal)

In summary, here's what happened at each electrode:

<u>C</u><u>athode</u>

- H⁺ selectively discharged

- ionic half equation: 2H⁺ (aq) +2e⁻ → H₂ (g)

<u>Anode</u>

- Cl⁻ ions selectively discharged

- ionic half equation: 2Cl⁻ (aq) → Cl₂ (g) + 2e⁻

3 0

3 years ago

A reaction that combines simpler reactants to form a new compound is called a

uranmaximum [27]

Answer:

A reaction that combines simpler reactants to form a new compound is called a

<h2>Synthesis reaction.</h2>

6 0

3 years ago