The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
Explanation:
Flow of electrons is possible with the help of a conducting medium like metal wire.
A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.
If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.
Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
Answer:
It is C
Explanation:
It is Cutting paper because cutting paper doesn't alter the chemical composition of paper
Answer : (4) Chromatography
Explanation :
Chromatography : It is a separation technique of a mixture by passing it through a medium in which components travels at different rates.
There are many types of chromatography but this is paper chromatography.
Paper chromatography : It is used to separate the colored substances. In paper chromatography, water is the mobile phase and paper is the stationary phase. The mixture of components moves at different speeds through the stationary phase so that they can be separated.
In paper chromatography, several colors can be separated based on their solubility. The more soluble a color is, the more readily it will dissolve in mobile phase and farther it will travel.
Answer:
4.07L of a 0.110M NaF are needed
Explanation:
Based on the reaction:
SrCl₂(aq)+2NaF(aq)⟶SrF₂(s)+2NaCl(aq)
<em>1 mole of strontium chloride react with 2 moles of NaF</em>
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361mL of 0.620M SrCl₂ solution has:
0.361L ₓ (0.620mol / L) = 0.22382 moles SrCl₂.
Moles of NaF for a complete reaction must be:
0.22382 moles SrCl₂ ₓ (2 mol NaF / 1 mol SrCl₂) = <em>0.44764 moles of NaF</em>
If you have a solution of 0.110M NaF, the moles of NaF needed are:
0.44764 moles of NaF ₓ (1L / 0.110mol NaF) = <em>4.07L of a 0.110M NaF are needed</em>
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