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tatiyna
2 years ago
13

Is it correct what I have done ​

Chemistry
2 answers:
antiseptic1488 [7]2 years ago
7 0

Anss\zyew:yes,the answer is very normal and well done

hram777 [196]2 years ago
5 0

Yeah it's correct...

Nice♥♥

Thanks♥♥

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20 points to anybody that can answer all 3 of these questions
Brilliant_brown [7]

Answer:

1. 2AgO2 --> 4 Ag + O2

2. Decomposition

3. 2NaBr + CaF2 --> 2NaF + CaBr2

Explanation:

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2 years ago
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The skeletal structure of Arachidonic acid, an essential fatty acid, is provided below. Arachidonic acid is best classified as:
Alexandra [31]

Skeletal structure is missing, so i have attached it.

Answer:

Option B: Omega 6 fatty acid

Explanation:

Arachidonic Acid is defined as an essential and unsaturated fatty acid. It's usually found in animal and human fat including the liver, brain, and also glandular organs. While in animals, it is a constituent of their phosphatides.

Arachidonic Acid is formed as a result of the synthesis from dietary linoleic acid and it's a preliminary stage in the biosynthesis of prostaglandins, thromboxanes, and leukotrienes.

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4 0
3 years ago
Where do electromagnetic waves move slowest?
Ludmilka [50]

Answer:

Solids

Explanation:

Electromagnetic waves travel fastest in empty space. But they are slowest in solids

8 0
2 years ago
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Identify the gas that has a root mean square velocity of 412 m/s at 191 K (potassium)
AVprozaik [17]
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5 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
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