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3 years ago

Is the dissolution reaction a spontaneous or non spontaneous?

1 answer:

6 0

Bottom line Your dissolution reaction is endothermic, but because dissolution reactions lead to states of higher entropy, your particular reaction is spontaneous (it occurs) at your specific temperature nonetheless

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What is the ph of a 0.014 M HCL solution?

Allushta [10]

Answer:

pH $= 1.853$

Explanation:

For every mole of hydrochloric acid, one mole of hydronium ion is required. Thus, in order to neutralize 0.014 moles of HCL, 0.014 moles of hydronium is required.

$[H_3O^+] = [HCl] = 0.014$

pH $= -log [H^+] = -log [H_3O^+]$

Substituting the available values in above equation, we can say that the pH of the solution is equal to

$- log (0.014)$

pH $= 1.853$

pH of a $0.014$ M HCL solution $= 1.853$

8 0

3 years ago

A student calculated that the amount of NaCl that should form in the lab (theoretical yield) is 2.50 grams. The actual amount fo

Kitty [74]

0.92%. student calculated that the amount of NaCl that should form in the lab

3 0

3 years ago

In order for a solute to dissolve in a silver, what must be true?

Elis [28]

It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium.

The resulting solution is called a saturated solution.

Certain substances are soluble in all proportions with a given solvent, such as ethanol in water.

This property is known as miscibility.

Under various conditions, the equilibrium solubility can be exceeded to give a so-called supersaturated solution, which is metastable.

The solvent is often a solid, which can be a pure substance or a mixture.

The species that dissolves, the solute, can be a gas, another liquid, or a solid.

Solubilities range widely, from infinitely soluble such as ethanol in water, to poorly soluble, such as silver chloride in water.

The term insoluble is often applied to poorly soluble compounds, though strictly speaking there are very few cases where there is absolutely no material dissolved.

The process of dissolving, called dissolution, is relatively straightforward for covalent substances such as ethanol.

When ethanol dissolves in water, the ethanol molecules remain intact but form new hydrogen bonds with the water.

When, however, an ionic compound such as sodium chloride (NaCl) dissolves in water, the sodium chloride lattice dissociates into separate ions which are solvated (wrapped) with a coating of water molecules.

Nonetheless, NaCl is said to dissolve in water, because evaporation of the solvent returns crystalline

7 0

2 years ago

If a straight-chain hydrocarbon is a gas at room temperature, how many carbon atoms will it have?

Orlov [11]

The fewer the carbon atoms, the closer it is to being a gas. The only one you have to check out is A which is hexane. You know that gasoline at the pumps has 8 carbons and its a liquid. So B and C are both not gases because they are above 8.

C6 (hexane) is a liquid at room temperature not a gas.

The answer is D. If there is a gas present, it must be C3

6 0

2 years ago

A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.

n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) = (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1 = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

Ea = (R*ln (k2/k1)) / ((1/T1)- (1/T2))

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1 - 1/ T3 = 1/400 - 1 /425 = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

4 0

3 years ago