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AURORKA [14]
2 years ago
14

According to Le Chatelier's principle, what happens to the equilibrium constant (K) when the concentration of the reactants is d

oubled? The value of the equilibrium constant (K) is doubled. The value of the equilibrium constant (K) is halved. The value of the equilibrium constant (K) remains the same. The value of the equilibrium constant (K) changes unpredictably.
Chemistry
2 answers:
slavikrds [6]2 years ago
5 0

Answer:

The value of the equilibrium constant (K) remains the same.

Explanation:

A state of dynamic equilibrium is said to have been achieved in a reaction system when the rate of forward reaction equals the rate of reverse reaction.

At equilibrium, doubling the initial concentration of reactants have no effect on the equilibrium constant K. The equilibrium will rather shift to the left or right as required in order to annul the constraint.

a_sh-v [17]2 years ago
4 0

Answer:

C

Explanation:

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8. Which of the following have bacteria in their root nodules that fix nitrogen?
Natalka [10]

Answer:

Option A

Explanation:

Leguminous plants like pulses etc. have root nodules comprising of rhizobacterium which live in a symbiotic relationship with the roots of the plant and in turn fix the nitrogen in the soil in the roots of the leguminous plants.

Hence, option A is correct

8 0
2 years ago
When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6
Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

Explanation:

Wavelength of the photon emitted = \lambda =646.3 nm =646.3\times 10^{-9} m

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

E=\frac{hc}{\lambda }

h = Planck's constant = 6.626\tiomes 10^{-34} Js

c = Speed of the light = 3\times 10^8 m/s

E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}

E=3.07\times 10^{-19} J

The energy difference between these 2p and 2s orbitals is 3.07\times 10^{-19} J

3 0
3 years ago
Describe the process by which ag+ ions are precipitated out of solution
MA_775_DIABLO [31]
Describe the process by which Ag+ ions are precipitated out of solution. 4. In your testing, several precipitates are formed, and then dissolved as complexes.
3 0
3 years ago
Read 2 more answers
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
HCl(?) + H2O(?) → H3O+(?) + Cl-(?)
larisa86 [58]

Answer:

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2 years ago
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