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Alexeev081 [22]
3 years ago
8

Is melting ice an endothermic or exothermic? its a big class discussion in chemistry and the teacher won't tell us until the who

le class agrees
Chemistry
1 answer:
Masteriza [31]3 years ago
3 0
It is an Endothermic because the ice cube is absorbing the heat making the ice cube melt. It can not be Endothermic because it is not releasing any heat it is melting.
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3 years ago
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The mass composition of a compound that assists in the coagulation of blood is 76.71% carbon, 7.02% hydrogen, and 16.27% nitroge
brilliants [131]

Answer : The empirical of the compound is, C_{11}H_{12}N_2

Explanation :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 76.71 g

Mass of H = 7.02 g

Mass of N = 16.27 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{76.71g}{12g/mole}=6.39moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.02g}{1g/mole}=7.02moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.27g}{14g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{6.39}{1.16}=5.5

For H = \frac{7.02}{1.16}=6.0\approx 6

For N = \frac{1.16}{1.16}=1

The ratio of C : H : N = 5.5 : 6 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : N = 11 : 12 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_{11}H_{12}N_2

Therefore, the empirical of the compound is, C_{11}H_{12}N_2

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The standard free-energy changes for the reactions below are given.Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/molATP → ADP
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Answer:

Gibbs free-energy of the reaction = (–12.5 kJ/mol)

Explanation:

The Gibbs free-energy of a reaction predicts the spontaneity or feasibility of a given chemical reaction.

<u>Given the standard Gibbs free energy changes</u>:

Phosphocreatine → creatine + Pi,  ∆G° = –43.0 kJ/mol     ...(1)

ATP → ADP + Pi , ∆G° = –30.5 kJ/mol      ....(2)

<u>Now to calculate the Gibbs free-energy of the given chemical reaction</u>: Phosphocreatine + ADP → creatine + ATP; the <em>equation (2) is reversed</em> to give:

ADP + Pi  → ATP, ∆G° = + 30.5 kJ/mol      ...(3)

<u>Now the equation (3) and (1) are added</u>, to give:

Phosphocreatine + ADP + Pi→ creatine + ATP + Pi

⇒ Phosphocreatine + ADP → creatine + ATP  

 

Therefore, to <u>calculate the Gibbs free-energy of the reaction, the standard Gibbs free energy changes of the equations (1) and (3) are added similarly</u>:

Gibbs free-energy of the reaction: ∆G° = (–43.0 kJ/mol) + ( + 30.5 kJ/mol) = (–12.5 kJ/mol)

<u><em>Therefore, the Gibbs free-energy of the reaction </em></u><u><em>= </em></u><u><em>(–12.5 kJ/mol)</em></u>

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6th grade science i mark as brainliest​
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