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2 years ago

"which of these nitrogen compounds is in the most reduced state?"

Chemistry

1 answer:

gtnhenbr [62]2 years ago

8 0

Because I can't see your options, all I can say is to look for the one whose numbers can't be simplified any further

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For double-helix formation, change in Gibbs free energy, ΔG, can be measured to be −54 kJ⋅mol−1 (−13 kcal⋅mol−1) at pH 7.0 in 1

77julia77 [94]

Answer:

Explanation:

Entropy change in the system : --

ΔG = −54 kJ⋅mol−1 (−13 kcal⋅mol−1) = −54 kJ⋅mol−1 (−13 x 4.2 kJ⋅mol−1)

= - 108.6 KJ / mol

ΔH = -251 kJ/mol (-60 kcal/mol) = -251 kJ/mol (-60 x 4.2 kJ/mol)

= - 503 KJ / mol

ΔG = ΔH - TΔS

ΔS = ( ΔH - ΔG ) / T

= - 503 + 108.6 / ( 273 + 25 ) KJ / mol k⁻¹

= - 1323.48 J / mol k⁻¹

Entropy change in the surrounding

+ 1323.48 J / mol k⁻¹

7 0

2 years ago

The energy difference between the 5d and 6s sublevels in gold accounts for its color. Assuming this energy difference is about 2

sergejj [24]

Answer:

$\lambda=459.1\times 10^{-7}\ m$ = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

Explanation:

Given that:- Energy = 2.7 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.602 × 10⁻¹⁹ J

So, Energy = $2.7\times 1.602\times 10^{-19}\ J=4.33\times 10^{-19}\ J$

Considering:-

$E=\frac{h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light

So,

$4.33\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$4.33\times \:10^{26}\times \lambda=1.99\times 10^{20}$

$\lambda=459.1\times 10^{-7}\ m$ = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

7 0

2 years ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

bekas [8.4K]

The correct values I believe would be a=1 b=-2 and c=-3.

8 0

2 years ago

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl

ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

$AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}{diluted}]$ = 0.0222 M

$[Cl^{-}{concentrated}]$ = 2.22 M

Now put all the given values in the above equation, we get:

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V

4 0

2 years ago

Solid aluminum metal and diatomic chlorine gas react spontaneously to form a solid product. Give the balanced chemical equation

ValentinkaMS [17]

When solid aluminum metal is reacted with diatomic chlorine gas, solid aluminum chloride is formed. This reaction is an example of synthesis or chemical combination in which two elements, aluminum and chlorine combine to form a new compound aluminum chloride.

Word equation: Aluminum (s)+ Chlorine (g)---> Aluminum chloride(s)

Molecular formula of the product formed is $AlCl_{3}$ .

Therefore the balanced chemical equation representing the reaction of solid aluminum with gaseous dichlorine can be represented as,

$2Al(s) + 3Cl_{2}(g)-->2AlCl_{3}(s)$

8 0

2 years ago