Answer:
b. Na+ leaves the salt bridge and enters enters the cathode
Explanation:
A galvanic cell or electrochemical cell depicts an oxidation -reduction half reactions (redox) reaction. it consists of two half cells ; one for the reduction reaction which involves the gain of electrons and the other for the oxidation reaction which involves the loss of electrons. One half cell contains the anode and oxidation occurs at the anode while the other half cell contains the cathode and reduction occurs at the cathode. The anode is usually connected to the cathode, a salt bridge is added to complete the circuit and allow current to flow. The salt bridge serves as a counter ions, they do not interfere with the electrochemical reaction but provides a passage for the migration of ions thereby preventing the cells from reaching equilibrium too quickly and thus the electrons in the salt are able to move along with any electrons.
In this galvanic cell, Cu at the anode losses two electrons to become Cu2+, and the electrons moves from the anode to the cathode where Mg2+ gain these two electrons to become negatively charged. Positively charged ions in the salt brigde Na+ will move to the cathode to pick negatively charged ions from the cathode solution. this helps to remove the strong negative charge from the cathode and allows the electrons to continue to move to the cathode.
In a study of the effect of lawn fertilizer on the health of a backyard pond, biologists grow too colonies of duckweed in a large flask to represent pond ecosystems.
Answer:
16.92 g C13H18O2
Explanation:
If the molecular mass of C13H18O2 is 206.3 amu, then molar mass of C13H18O2 is 206. 3 g/mol.
So, if we have 0.082 mol C13H18O2, than it is
206.3 g/mol*0.082 mol = 16.92 g C13H18O2.
Answer:
Mass percentage → 0.074 %
[F⁻] = 741 ppm
Explanation:
Aqueous solution of flouride → [F⁻] = 0.0390 M
It means that in 1L of solution, we have 0.0390 moles of F⁻
We need the mass of solution and the mass of 0.0390 moles of F⁻
Mass of solution can be determined by density:
1g/mL = Mass of solution / 1000 mL
Note: 1L = 1000mL
Mass of solution: 1000 g
Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g
Mass percentage → (Mass of solute / Mass of solution) . 100
(0.741 g / 1000 g) . 100 = 0.074 %
Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)
0.741 g . 1000 mg/1g = 741 mg
1000 g . 1 kg/1000 g = 1kg
741 mg/1kg = 741 ppm