A(n) 7.88-mol sample of carbon monoxide was stored in a 30.0-L container at 49.4°C. What is the

8. How much enthalpy/heat is transferred when 0.5113

Gennadij [26K]

First convert to moles:
0.5113 g / 17 g/mol = 0.0301 mol

Now create a ratio based on the reaction provided to solve for the unknown:

4 NH3 / -905.4 kJ = 0.0301 mol NH3 / x kJ
x = -6.808 kJ

8 0

3 years ago

Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0

3 years ago

As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit

jeka94

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced 0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles = 0.0003663 moles

Mass of K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

7 0

4 years ago

How do scientists measure the idea of time so long ago?

professor190 [17]

The Ancient Egyptians used simple sundials and divided days into smaller parts, and it has been suggested that as early as 1,500BC, they divided the interval between sunrise and sunset into 12 parts. ... Known as a clepsydra, it uses a flow of water to measure time.

6 0

3 years ago

Given 2.0 moles of nitrogen and plenty of hydrogen, how many moles of NH3 are formed?

kolezko [41]

3.0 moles

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3 0

3 years ago