Answer:
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In a galvanic cell, the flow of electrons will be from the anode to cathode through the circuit .
Whether a cell is an electrolysis cell (non-spontaneous chemistry driven by forcing electricity from an external energy source) or a galvanic cell (spontaneous chemistry driving electricity), will determine the charge of the anode and the cathode. Depending on where the electrons encounter resistance and find it difficult to pass, a negative charge may emerge. Therefore, you cannot determine the direction of the current just on the charge on the electrode.
Oxidation and reduction always take place at the anode and cathode, respectively.
An element undergoes oxidation when it surrenders one or more electrons to become more positively charged. These electrons leave the chemicals in any type of cell and travel to the anode, where they enter the external circuit.
An element picks up an electron during reduction to become more negatively charged (less positive, lower oxidation state). These electrons are captured from the external circuit at the cathode in both types of cells.
Therefore, no matter what kind of cell you are dealing with, the oxidizing chemicals at the anode transfer the electrons to the external circuit; these electrons then move through the circuit from the anode to the cathode, where they are captured by the reducing chemicals. The electrons always go from the anode to the cathode via the external circuit.
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Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:
If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):
6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
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Answer:
it is the primary electricity in solid. they also make up an atom.
Explanation:
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Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
