Answer:
Option A. 2.82 L
Explanation:
Step 1:
Data obtained from the question.
pH = 13.55
Step 2:
Determination of the pOH of the solution.. This is illustrated below:
pH + pOH = 14
pH = 13.55
13.55 + pOH = 14
Collect like terms
pOH = 14 - 13.55
pOH = 0.45
Step 3:
Determination of the concentration of the OH ion.
This is illustrated below:
pOH = - Log [OH-]
pOH = 0.45
0.45 = - Log [OH-]
- 0.45 = Log [OH-]
[OH-] = antilog (- 0.45)
[OH-] = 0.355 M
Step 4:
Determination of the molarity of KOH. This is illustrated below:
First, we'll write the dissociation equation of KOH as follow:
KOH —> K+(aq) + OH-(aq)
From the balanced equation above,
1 mole of KOH produced 1 mole of OH-.
Therefore, 0.355 M of KOH will definitely produce 0.355 M of OH-.
The molarity of KOH is 0.355 M
Step 5:
Determination of the volume of the solution needed to dissolve 1 mole of KOH. This is illustrated below:
Mole of KOH = 1 mole
Molarity of KOH = 0.355 M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 1/0.355 M
Volume = 2.82L
Number of moles is defined as the ratio of given mass in grams to the molar mass of compound.
Number of moles =
Now, put the value of given mass of in grams and molar mass of in g/mol i.e. 13 g/mol.
Thus,
number of moles =
= 0.692 mol
Hence, number of moles of = 0.692 mol
Chromatography is a pretty accurate description of what happens to ink on wet paper, because it literally means "color writing" (from the Greek words chroma and graphe). Really, though, it's a bit of a misnomer because it often doesn't involve color, paper, ink, or writing. Chromatography is actually a way of separating out a mixture of chemicals, which are in gas or liquid form, by letting them creep slowly past another substance, which is typically a liquid or solid. So, with the ink and paper trick for example, we have a liquid (the ink) dissolved in water or another solvent creeping over the surface of a solid (the paper).
The essential thing about chromatography is that we have some mixture in one state of matter (something like a gas or liquid) moving over the surface of something else in another state of matter (a liquid or solid) that stays where it is. The moving substance is called the mobile phase and the substance that stays put is the stationary phase. As the mobile phase moves, it separates out into its components on the stationary phase. We can then identify them one by one.
C.Only CuS will precipitate from solution
Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together.
Na₂S(aq) + CuCl₂(aq) → Products
Na₂S(aq) + CuCl₂(aq) → CuS(s) ↓ + 2NaCl(aq)
When the sulfide bond to the cation Cu²⁺ it makes a precipitate (s)
Thus,the correct answer is C.
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Answer:
The correct state for the answer is c.
Only CuS will precipitate from solution
Explanation:
We analyse the compounds for the reaction and we write the equation:
Na₂S(aq) + CuCl₂(aq) → Products
For this case, the products are NaCl and CuS. The complete balanced reaction is:
Na₂S(aq) + CuCl₂(aq) → CuS(s) ↓ + 2NaCl(aq)
When the sulfide bond to the cation Cu²⁺ it makes a precipitate (s)
Salts from chlorides are soluble except for the Ag⁺, Pb⁺ or Cu⁺
Salts from S⁻² which are soluble, are found in the group 2 of the Periodic Table (Ca²⁺, Ba²⁺, Mg²⁺)
The correct state for the answer is c.
How many bonding electrons are present in this compound?
How many lone pair (non-bonding) electrons are present in this compound?
Answer:
Valence electrons in XeCl2 = 8 + 7 + 7 = 22.
Bonding electrons = 4.
Nonbonding electrons = 18.
Explanation:
Hello.
In this case, you can see the Lewis structure on the attached picture, in which you can see that there are since xenon has 8 valance electrons and each chlorine has 7 valence electrons, the total amount of valence electrons is:
Valence electrons in XeCl2 = 8 + 7 + 7 = 22.
Moreover, since each chlorine atom is bonding with one of the eight electrons of xenon (Lewis structure), we can see there are 4 bonding electrons.
Finally, since there are six nonbonding electrons per chlorine atom and six nonbonding electrons in xenon, the overall nonbonding electrons are:
Nonbonding electrons in XeCl2 = 6 + 6 + 6 = 18.
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When Americium (Am-241) undergoes alpha decay(He-4) it forms neptunium (Np-237) based on the following pathway:
²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂
The energy released in given as:
ΔE = Δmc²
where Δm = mass of products - mass of reactants
= [m(Np-237) + m(He-4)] - [m(Am-241)]
= 237.0482+4.0015-241.0568 = -0.0073 g/mol = -7.3 * 10⁻⁶ kg/mol
ΔE = -7.3*10⁻⁶ kg/mol * (3*10⁸ m/s)² = -5.84*10¹¹ J/mol