Answer:
V = 0.24 L.
Explanation:
Hello there!
In this case, it is possible to approach this problem by using the ideal gas equation defined by:
Whereas our unknown is V, volume, and we solve for it as follows:
Then since the STP conditions are ate 273.15 K and 1 atm, the volume of 0.011 moles of CO2 gas turns out to be:
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Answer: B
Explanation: In thermodynamics, the Gibbs free energy (IUPAC recommended name: Gibbs energy or Gibbs function; also known as free enthalpy to distinguish it from Helmholtz free energy) is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system.
Answer:
The sun is the large astronomical body that is located at the center of the solar system. The interior structure of the sun are as follows-
- Core- It represents the extreme interior portion of the sun. This is the hottest region of the sun and there occurs the process of constant nuclear fusion reactions that fuel the sun and helps in burning. Here the nucleus of two hydrogen atoms is fused together to form a heavy nucleus of a helium atom. It has a thickness of about 500 km.
- Radiative zone- This is the layer that surrounds the core of the sun. From this layer, the energy is radiated outward where the protons carry these energies in the form of thermal radiation. Here, the process of radiation takes place.
- Convective zone- This is the outer layer of the sun's interior. It is about 200,000 km thick, and in this layer, the energy constantly flows, allowing the heat to move upward by undergoing the process of convection.
Answer:
Right
Explanation:
The given parameters are;
The soluble sulfate formed by the metal, M = M₂SO₄
In the galvanic cell formed by the metal M, we have;
The concentration of M₂SO₄ in the left half cell = 50.0 mM = 0.05 M
The concentration of M₂SO₄ in the right half cell = 5.00 M
In the galvanic cell, the metal 'M' will be dissolved into the solution with lower concentration as M²⁺ which is the left half cell, making the cell negative and the solution more concentrated
In the right half cell, the metal 'M²⁺' in the solution will be plated unto the electrode making the solution less concentrated and the electrode in the right half cell will be the positive electrode
Therefore;
The electrode which will be positive is the electrode in the right half cell.