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2 years ago

Which particle is gained and lost during a redox process?

Chemistry

1 answer:

alekssr [168]2 years ago

6 0

Answer:

electrons

Explanation:

A redox reaction is one in which oxidation and reduction takes place. in the sense of

oxidation being the loss of electrons and reduction the gain of electrons.

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Write the Arrhenius Base reaction for the following: Sr(OH)2

igor_vitrenko [27]

Answer:

Explanation:

Sr(OH)2 (aq) ⇔ Sr+2 (aq) + 2OH- (aq)

8 0

3 years ago

The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

7 0

3 years ago

When the following equation is balanced, what is the coefficient of oxygen?

Gwar [14]

Answer:

The answer is 3

C2H5OH + O2 CO2 +H2O (unbalanced)

C2H5OH +3O2(g). 2CO2(g)+3H2O(balanced)

4 0

3 years ago

The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 104 (?-m)-1. The electron concentration is kno

Julli [10]

Answer:

7.43 × 10²⁴ m⁻³

Explanation:

Data provided in the question:

Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹

Electron concentration, n = 2.9 × 10²² m⁻³

Electron mobility, $\mu_n$ = 0.14 m²/V-s

Hole mobility, $\mu_p$ = 0.023 m²/V-s

Now,

σ = $nq\mu_n+pq\mu_p$

σ = $q(n\mu_n+p\mu_p)$

here,

q is the charge on electron = 1.6 × 10⁻¹⁹ C

p is the hole density

thus,

2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )

1.75 × 10²³ = 0.406 × 10²² + 0.023p

17.094 × 10²² = 0.023p

p = 743.217 × 10²²

p = 7.43 × 10²⁴ m⁻³

5 0

4 years ago

What volume of 3.00 MM HClHCl in liters is needed to react completely (with nothing left over) with 0.750 LL of 0.500 MM Na2CO3N

AlladinOne [14]

Answer: The volume of HCl needed is 0.250 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For sodium carbonate:

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

$0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol$

The chemical equation for the reaction of sodium carbonate and HCl follows:

$Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3$

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = $\frac{2}{1}\times 0.375=0.750mol$ of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

$3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L$

Hence, the volume of HCl needed is 0.250 L

8 0

3 years ago