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mars1129 [50]
2 years ago
6

How to calculate magnitude of k

Chemistry
1 answer:
olasank [31]2 years ago
6 0
Use Arrhenius equation: 
k = A*exp(-Ea/RT) 
We have: 
1.35x10^2/s = A*exp(-85600/(8.314*298.15)) 
or: A = 1.342x10^17/s 
It is a piece of cake to calculate: 
k = 1.342x10^17*exp(-85600/(8.314*348.15)) 
= 1.92x10^4/s
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Why is it that the decomposition of nitroglycerin can result in so
patriot [66]

Answer:

with the molecular formula C3H5(ONO2)3, has a high nitrogen content (18.5 percent) and contains sufficient oxygen atoms to oxidize the carbon and hydrogen atoms while nitrogen is being liberated, so that it is one of the most powerful explosives known.

Explanation:

NTG reduces preload via venous dilation, and achieves modest afterload reduction via arterial dilation. These effects result in decreased myocardial oxygen demand. In addition, NTG induces coronary vasodilation, thereby increasing oxygen delivery.

3 0
3 years ago
1. The s orbitals are not symmetrical in shape.<br> a. TRUE<br> O. FALSE
ira [324]

The s orbitals are not symmetrical in shape is a FALSE statement.

An s orbital is so symmetric, more specifically spherically symmetric that it looks the same from all directions.

  • The atomic orbitals in the atoms of elements differ in shape.

In essence, the electrons they describe have varying probability distributions around the nucleus. The spherical symmetry of s orbitals is evident in the fact that all orbitals of a given shell in the hydrogen atom have the same energy.

  • All s orbitals are spherically symmetrical. Put simply, an electron that occupies an s orbital can be found with the same probability at any orientation (at a distance) from the nucleus.

The s orbitals are therefore represented by a spherical boundary surface which is a surface which captures a high proportion of the electron density.

Read more:

brainly.com/question/5087295

4 0
1 year ago
Adjustment to environmental conditions is known as
alukav5142 [94]

Answer:

A.

Explanation:

Adjust = Adapt if that makes sense.

8 0
2 years ago
Read 2 more answers
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
2 years ago
Se construye una pila galvánica con una barra de cobre sumergida en una disolución 1M de cationes Fe+2 y una barra de plata sume
pashok25 [27]

Answer:

El potencial celular estándar, E_{cell} is +0.46 V

Explanation:

Las reacciones de media célula son;

Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V

Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V

Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V

y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V

 que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo

Por lo tanto, en el ánodo, tendremos

Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)

En el cátodo

2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)

E_{cell} = E_c + E_a = -0.34 + 0.8 = +0.46 \, V

El potencial celular estándar, E_{cell} = +0.46 V

4 0
3 years ago
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