The addictive power of nicotine is enhanced by which of the following chemicals?

Using the following thermochemical data, what is the change in enthalpy for the following reaction?

slavikrds [6]

Answer:

D. -120.9 kJ

Explanation:

According to Hess's law ,the total enthalpy change for a reaction is the sum of all changes regardless of the stages or the steps of the reaction.

$CaO + 2HCl \rightarrow CaCl_{2} + H_{2}O\ (\Delta H = -186\ kJ)$ ....(1)

$CaO + H_{2}O\rightarrow Ca(OH)_{2}\ (\Delta H = - 65 \ kJ)$

(this reaction should be reversed in order to reach the required reaction )

On reversing the reaction the sign of $\Delta H$ get reversed.

(In this case change sign from '-' to'+'. Hence $\Delta H$ = + 65 kJ)

$CaO + 2HCl \rightarrow CaCl_{2} + H_{2}O\ (\Delta H = - 186\ kJ)$ ....(1)

$Ca(OH)_{2} \rightarrow CaO + H_{2}O\ (\Delta H = + 65 kJ )$ ......(2)

Adding equation (1) and (2)

$Ca(OH)_{2} + 2HCl \rightarrow CaCl_{2} + 2H_{2}O[tex][tex]Delta H = - 186 + 65 = - 121\kJ$

$Delta H = - 121\kJ$ (It is nearly equal to -120.9 kJ)

7 0

3 years ago

How many molecules would there be in 10.5 L of carbon dioxide at 40.0 C and 252 kPa

kakasveta [241]

Answer:

6.14×10²³ molecules

Explanation:

Data obtained from the question include:

Volume (V) = 10.5L

Temperature (T) = 40°C

Pressure (P) = 252 kPa

Next, we shall determine the number of mole of CO2 present.

This can be obtained by using the ideal gas equation:

PV = nRT

Volume (V) = 10.5L

Temperature (T) = 40°C = 40°C + 273 = 313K

Pressure (P) = 252 kPa

Gas constant (R) = 8.31 KPa.L/Kmol

Number of mole (n) =.?

PV = nRT

252 x 10.5 = n x 8.31 x 313

Divide both side by 8.31 x 313

n = (252 x 10.5) /(8.31 x 313)

n = 1.02 mole

Therefore, 1.02 mol of CO2 is present.

Now, we can obtain the number of molecules of CO2 present as follow:

From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules.

This means that 1 mole of CO2 also contains 6.02×10²³ molecules.

Now, if 1 mole of CO2 contains 6.02×10²³ molecules,

Then 1.02 mole will contain = 1.02 x 6.02×10²³ = 6.14×10²³ molecules.

Therefore, 6.14×10²³ molecules of CO2 is present.

8 0

4 years ago

Calculate the ΔH o for the reaction: Fe3+(aq) + 3OH−(aq) → Fe(OH)3(s) ΔH o = kJ/mol Substance ΔH o f (kJ/mol) ΔG o f (kJ/mol) S

timofeeve [1]

Answer:

ΔHr = -86.73 kJ/mol

Explanation:

Using Hess's law, you can calculate ΔH of any reaction using ΔH°f of products and reactants involed in the reaction.

Hess law: ∑nΔH°f products - ∑nΔH°f reactants = ΔHr

-Where n are moles of reaction-

For the reaction:

Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)

Hess law is:

ΔHr = ΔH°f Fe(OH)₃ - ΔH°f Fe³⁺ - 3×ΔH°f OH⁻

Where:

ΔH°f Fe(OH)₃: −824.25 kJ/mol

ΔH°f Fe³⁺: −47.7 kJ/mol

ΔH°f OH⁻: −229.94 kJ/mol

Replacing:

ΔHr = −824.25 kJ/mol - (−47.7 kJ/mol) - (3×-229.94 kJ/mol)

ΔHr = -86.73 kJ/mol

8 0

3 years ago

__________ may be shared by or transferred to other atoms.

Vsevolod [243]

D.) "Electrons" may be shared by or transferred to other atoms.

Hope this helps!

8 0

3 years ago

Read 2 more answers

Potassium-40 is a radioactive isotope that decays into a single argon-40 atom and other particles with a half-life of 1:25 billi

stiks02 [169]

Answer:

0.147 billion years = 147.35 million years.

Explanation:

It is known that the decay of a radioactive isotope isotope obeys first order kinetics.

Half-life time is the time needed for the reactants to be in its half concentration.

If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

The half-life of Potassium-40 is 1.25 billion years.

For, first order reactions:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.

Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

At the time needed to be determined:

8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.

If we start with 100% Potassium-40:

∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.

8 0

3 years ago