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sp2606 [1]
3 years ago
10

Practice with Dilution Calculations IV,

Chemistry
1 answer:
larisa [96]3 years ago
7 0

Answer:

THE INITIAL VOLUME OF THE STOCK SOLUTION IS 0.0392 L OR 39.2 mL

Explanation:

To calculate the volume of a solution before dilution, we make use of the dilution formula;

C1 V1 = C2 V2

C1 = the concentration of the stock solution = 15 M

V 1 = the volume of stock solution = ?

C2 = concentartion of final solution = 2.35 M

V2 = volume of final solution = 250 mL = 250/ 1000 = 0.25 L

Rearranging the formula, we obtain;

V1 = C2V2 / C1

V 1 = 2.35 * 0.25 / 15

V1 = 0.5875 / 15

V1 = 0.0392 L

the initial volume of the stock solution is 0.0392 L or 39.2 mL

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The salt that is formed during the reaction between potassium hydroxide and hydrochloric acid is
liubo4ka [24]

Answer:

Sodium chloride (NaCl)

Explanation:

According to Arrhenius, an acid is a substance that dissociates to give protons, whereas a base dissociates to give hydroxide ions in an aqueous solution.

Therefore, a chemical reaction in which an Arrhenius acid reacts with an Arrhenius base to give salt and water, is known as a Neutralization reaction.

For example: <u>Neutralization reaction of hydrochloric acid (HCl) and sodium hydroxide (NaOH)</u>

A strong acid, hydrochloric acid (HCl) reacts with a strong base, sodium hydroxide (NaOH) to give salt, sodium chloride (NaCl) and water (H₂O).

<em>The chemical equation for this Neutralization reaction:</em>

HCl (acid) + NaOH (Base) → NaCl (Salt) + H₂O (Water)

<u />

<u>Therefore, </u><u>sodium chloride (NaCl) is the salt formed</u><u> during the chemical reaction of hydrochloric acid (HCl) and sodium hydroxide (NaOH).</u>

3 0
3 years ago
To gravimetrically analyze the silver content of a piece of jewelry made from an alloy of Ag and Cu, a student dissolves a small
Ipatiy [6.2K]

Answer:

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions.

Explanation:

The student have in solution Ag⁺ and Cu²⁺ ions but he just want to analyze the silver, that means he need to separate ions.

Centrifuging the solution to isolate the heavier ions <em>FALSE </em>Centrifugation allows the separation of a suspension but Ag⁺ and Cu²⁺ are both soluble in water.

Adding enough base solution to bring the pH up to 7.0 <em>FALSE </em>At pH = 7,0 these ions are soluble in water and its separation will not be possible.

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions <em>TRUE </em>For example, the addition of Cl⁻ will precipitate the Ag⁺ as AgCl(s) allowing its separation.

Evaporating the solution to recover the dissolved nitrates. <em>FALSE</em> . Thus, you will obtain the nitrates of these ions but will be mixed doing impossible its separation.

I hope it helps!

8 0
3 years ago
Given these reactions, where X represents a generic metal or metalloid.
AlexFokin [52]
Number 1 is the correct answer
4 0
4 years ago
For a gaseous carbon atom in its' ground state which of the following are possible sets of quantum numbers for two different ele
MAXImum [283]

Answer:

Option e=>

electron 1; n l ml ms = 2 1 0 +1/2.

electron 2;n l ml ms= 2 1 -1 +1/2.

Explanation:

The electronic configuration of carbon in its ground state is 1s2 2s2 2p2. The excited electronic configuration of carbon is given as; 1s2 2s1 2p3 because one of the electrons moves to the 2p sub- orbital because of the absorption of the atom.

Recall that there are four different types of quantum numbers and they are the principal quantum number,n; the Azimuthal quantum number, l; the magnetic quantum number, ml and the spin quantum number, Ms.

Our principal quantum number,n is TWO(2) which is the highest energy level from the electronic configuration written above.

The Azimuthal quantum number is 1 from l= n - 1.

The magnetic quantum number is -1 and 0 because we have two electrons filling the 3 orbitals of 2p sublevels. That is;

2px = -1 , 2py = 0 and 2pz = +1.

Only 2px and 2py will be considered because we have 2 electrons on the 2p in the ground state electronic configuration.

Therefore, the first electron will have ml= 0 and the second electron will have ml= -1.

Spin quantum number, ms; in both electrons the spins is an up spin so we have a +1/2 for both electron.

Hence; electron 1; n l ml ms = 2 1 0 +1/2.

Electron 2;n l ml ms= 2 1 -1 +1/2.

8 0
3 years ago
EXPERIMENT 1: Identify the oxidation and reduction half-reactions that occur in
Hoochie [10]

Answer:

Cathode: Mn → Mn²⁺  +  2e⁻    (Oxidation)

Anode: Zn²⁺  +  2e⁻  →  Zn   (Reduction)

Mn | Mn²⁺ ||  Zn²⁺  | Zn

Explanation:

To identify the half reaction we need to see the oxidation states.

Mn(s) → Ground state → Oxidation state = 0

Mn(NO₃)₂ → Mn²⁺  → The oxidation state has increased.

This is the oxidation reaction. It has released two electrons:

Mn → Mn²⁺  +  2e⁻

Zn(NO₃)₂ →  Zn²⁺

Zn → Ground state → The oxidation state was decreased.

This is the reduction reaction. It has gained two electrons:

Zn²⁺  +  2e⁻  →  Zn

Cathode: Mn → Mn²⁺  +  2e⁻

Anode: Zn²⁺  +  2e⁻  →  Zn

Mn | Mn²⁺ ||  Zn²⁺  | Zn

4 0
3 years ago
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