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kap26 [50]
3 years ago
10

Calculate the mass in grams of iodine (i 2 ) that will react completely with 20.4 g of aluminum (al) to form aluminum iodide (al

i
Chemistry
1 answer:
garik1379 [7]3 years ago
8 0
3I₂ + 2Al → 2AlI₃

m(I₂)=3M(I₂)m(Al)/{2M(Al)}

m(I₂)=3*253.8*20.4/{2*27.0}=287.64 g
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rosijanka [135]

Answer:

A)P5010

Explanation:

Penta- means 5 and deca/deco- means 10

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Test anxiety symptoms can include
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Answer:

D.) All of the above

Explanation:

Anxiety can cause sweating, chaky hands, headaches etc

4 0
3 years ago
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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
Jamal is working with three ionic compounds: sodium chloride, calcium sulfide, and barium oxide. His teacher asks him which are
Lubov Fominskaja [6]

Answer:

option A is correct  ( sodium, calcium and barium)

Explanation:

Given compounds:

Sodium chloride , calcium sulfide, barium oxide

We know that metals form positive ions. In order to solve the problem we must identify the metals from given compounds.

Na⁺Cl⁻

Ca²⁺S²⁻

Ba²⁺O²⁻

We can see that sodium, calcium and barium contain positive charges.

Thus option A is correct.

Because sodium have one valance electron. When it combine with chlorine sodium lose its one electron to complete the octet and chlorine accept it to complete its octet. Thus sodium form positive ion and chlorine form negative ion.

Similarly barium and calcium are present in group 2. Both have two valance electron. When they lose them cation are formed.

Other option are incorrect because,

Option B have sulfur and oxygen which are anion.

Option C have chlorine which is also anion

Option D have chlorine, sulfur and oxygen that are anions.

4 0
3 years ago
A more reactive metal displaces a less reactive metal from salt solution. Such reactions are called:
motikmotik

Answer:

Displacement Reaction.

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