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sveta [45]
1 year ago
15

Amy performed an experiment in lab. She improperly mixed the chemicals, and an explosion of light, sound, and heat occurred. Whe

n Amy mixed the chemicals, energy was _______.
Chemistry
2 answers:
svlad2 [7]1 year ago
5 0

<span>The energy that was produced is called, heat combustion. It is an energy released in the form of heat when chemicals are mixed. An explosion of light and sound are the common characteristics of heat combustion. The chemical reaction takes place because of the presence of oxygen and hydrocarbon or organic molecule substances that when mixed form carbon dioxide and water which then releases heat and explosive characteristics.</span>

ElenaW [278]1 year ago
3 0

in other words the answer is transformed


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Answer:

¨it is negatively charged¨ i took the science test in edgeunity and got it right  

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Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this temperature. substance: ba2+(
jeka57 [31]

Answer : the correct answer for ksp = 1.59 * 10⁻⁹

Following are the steps to calculate the ksp of reaction

BaCO₃ →Ba ²⁺ + CO₃²⁻ :

Step 1 : To find ΔG° of reaction :

ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :

ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .

Given : ΔG° for Ba²⁺ ( product )= -560.7 \frac{KJ}{Mol}

ΔG° for CO₃²⁻ (product ) =- 528.1 \frac{KJ}{Mol}

ΔG° BaCO₃ ( reactant) = –1139 \frac{KJ}{Mol}

Plugging value in formula :

ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )

⁻ = ( -560.7 \frac{KJ}{Mol} + 528.1 \frac{KJ}{Mol} ) - ( -1139 \frac{KJ}{Mol} )

= ( -1088.8 \frac{KJ}{Mol}) - (-1139 \frac{KJ}{Mol} )

= - 1088.8 \frac{KJ}{Mol} + 1139 \frac{KJ}{Mol}

ΔG° of reaction = 50.2 \frac{KJ}{Mol}

Step 2: To calculate ksp from ΔG° of reaction .

The relation between Ksp and ΔG° is given as :

ΔG° = -RT ln ksp

Where ΔG° = Gibb's Free energy R = gas constant T = Temperature

Ksp = Solubility constant product .

Given : ΔG° of reaction = 50.2 \frac{KJ}{Mol}

T = 298 K R = 8.314 \frac{J}{Mol * K}

Plugging values in formula

50.2 \frac{KJ}{mol}  =  -  8.314 \frac{J}{mol * K} * 298 K * ln  ksp

50.2 \frac{KJ}{mol}  =  - 2477.572 \frac{J}{mol} * ln K

((Converting 2477 \frac{J}{mol}  to \frac{KJ}{mol}

Since , 1 KJ = 1000 J So , 2477 \frac{J}{mol}  * \frac{1 KJ}{1000J}  = 2.477 \frac{KJ}{mol} ))

Dividing both side by - 2.477 \frac{KJ}{mol}

\frac{50.2\frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} = \frac{-2.477 \frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} * ln ksp

ln ksp = ln ksp = -20.27 \frac{KJ}{mol}

Removing ln :

ksp = 1. 59 * 10⁻⁹

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2 years ago
The beta oxidation pathway degrades activated fatty acids (acyl-CoA) to acetyl-CoA, which then enters the citric acid cycle. Add
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Answer:

The correct statements are given below

Explanation:

b Enoyl CoA isomerase an enzyme that converts cis double bonds to trans double bonds in fatty acid metabolism,bypasses a step that reduces Q,resulting in the higher ATP yield.

c Even chain fatty acids are oxidized to acetyl CoA in the beta oxidation pathway.

f The final round of beta oxidation foe a 13 carbon saturated fatty acid yields acetyl CoA and propionyl CoA a three carbon fragment.

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Answer: the way root cells reproduce to increase root length

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Rudolf Ludwig Carl Virchowas known as the founder of social medicine and also the father of modern pathology.

Virchow posited that all cells are gotten from already existing cells and he used this in his work towards cellular pathology, as it was made clear that diseases takes place at the cellular level. He posited that the cells that are malfunctioning cause diseases.

Based on the above analysis, the image that Tumu would most likely use in his assignment to feature Rudolf Virchow is the way root cells reproduce to increase root length.

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Radon (Rn) is the heaviest and the only radioactive member of Group 8A(18), the noble gases. It is a product of the disintegrati
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Answer:

V=9.89x10^{-7}\frac{L}{day}

Explanation:

Hello,

On the attached document, you will find the procedure regarding the required volume. At first, it is necessary to know atoms by day produced, by using a rule of three. Subsequently, we compute the moles per day and finally the volume via the ideal gas equation at STP.

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