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Nonamiya [84]
2 years ago
13

For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

Chemistry
1 answer:
harina [27]2 years ago
6 0

Answer:

Q= 245 =2.5 * 10^2

Explanation:

ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK

R= 8,314 J/molK, T=298K

ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol

ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ

→ 5.5 = LnQ → Q= 245 =2.5 * 10^2

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Propose a plausible mechanism for the reaction f2 + 2clo2 → 2fclo2 given that the rate law for the reaction is rate = k[f2][clo2
shepuryov [24]

<u>The given reaction is:</u>

F2 + ClO2 → 2FClO2

Rate = k[F2][ClO2]

<u>Explanation:</u>

The possible mechanism for this reaction can be broken down into two steps with the slow step being the rate determining step

Step 1:       F2 + ClO2 → FClO2 + F ----------- Slow

Step 2:      F + ClO2 → FClO2           ----------- Fast

-----------------------------------------------------------

Overall:  F2 + 2ClO2 → 2FClO2

Rate = k[F2][ClO2]

 


8 0
1 year ago
Read 2 more answers
A few drops of a mixture of sodium hydroxide(NaOH) solution and copper (II) tetraoxosulphate (VI) solution were added to a sampl
Kaylis [27]

Answer:

It is a test for proteins in urine

Explanation:

The Biuret reagent is made of sodium hydroxide (NaOH) and hydrated copper(II) sulfate. The biuret reagent is commonly used to test for proteins. The biuret test is also known as Piotrowski's test. It is a chemical test commonly used in detecting the presence of peptide bonds. In the presence of peptides, a copper(II) ion forms purple-colored coordination complexes in an alkaline solution.

Hence the addition of this Biuret reagent to a urine sample in a test tube aims to detect the presence or absence of proteins in the given urine sample. If there is protein in the urine sample, the blue colour of the Biuret reagent turns purple. If there is no protein in the urine sample, the Biuret reagent remains blue.

In an alkaline solution, copper II is able to form a complex with the peptide bonds in proteins. Once this complex has been formed, the Biuret solution turns from a blue color to a purple color. This is the positive test for proteins.

5 0
1 year ago
A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the
Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047  mol/dm³

The concentration of the sulfuric acid is 0.047  mol/dm³

Download docx
7 0
1 year ago
Three structural isomers have the formula C5H12.C5H12. Draw and name the isomers using IUPAC names. Draw the isomer with five ca
chubhunter [2.5K]

Answer:

Explanation:

Answer in attached file .

8 0
2 years ago
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:(a) |(b) Sr(c) K(d) N(e) S(f)
Pavel [41]

Answer:

(a) I⁻ (charge 1-)

(b) Sr²⁺ (charge 2+)

(c) K⁺ (charge 1+)

(d) N³⁻ (charge 3-)

(e) S²⁻ (charge 2-)

(f) In³⁺ (charge 3+)

Explanation:

To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.

(a) |

I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).

(b) Sr

Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).

(c) K

K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).

(d) N

N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).

(e) S

S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).

(f) In

In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).

3 0
2 years ago
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