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Ratling [72]
2 years ago
12

25. Which is an irreversible process?

Chemistry
2 answers:
Nesterboy [21]2 years ago
3 0

The answer is 1) mixing of two gases by diffusion.

Once 2 gases are mixed by diffusion, they cannot be obtained from the mixture of gases. Hence, it is considered to be an irreversible process.

sergejj [24]2 years ago
3 0
3 dissolution of NaCl in water, since chemical changes are irreversible
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What is the correct name for MgF2?
maxonik [38]
MgF2 is <span>Magnesium fluoride

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4 years ago
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9. According to an "alternative theory", H2O is
Jlenok [28]

Answer:(4) ----accepts  a proton

Explanation:

H2O water can produce both hydrogen and   hydroxide ions

H2O --> H+ + OH-

According to the Bronsted-Lowry theory, it can be a proton donor and a proton acceptor.this means that It can donate a hydrogen ion to become its conjugate base, or  can accept a hydrogen ion to form its conjugate acid,

When , a water molecule, H2O accepts a proton it will act as a Brønsted-Lowry base especially when dissolved in a strong acidic medium. for eg

HCl + H2O(l) → H3O+(aq) + Cl−(aq)

Here, Hydrochloric acid is a strong acid and ionizes completely in  water, since it is more acidic than water, the water will act as a base.

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3 years ago
How many moles of potassium chloride are in 28 grams of KCl?
stiv31 [10]

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0.3758moles

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7 0
3 years ago
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Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of
bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol..[3]

NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

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How quickly must non-frozen ready-to-eat foods be consumed?
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Answer:

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