When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
Answer:
±0.005 g
Explanation:
The uncertainty depends on whether the measurement was obtained manually or digitally.
1. Manual
The minimum uncertainty is ±0.01 g.
It may be greater, depending on random or personal errors
2. Digital
Most measurements of mass are now made on digital scales.
A digital device must always round off the measurement it displays.
For example, if the display reads 20.00, the measurement must be between 20.005 and 19.995 (±0.005).
If the measured value were 20.006, the display would round up to 20.01.
If the measured value were 19.994, the display would round down to 19.99.
The uncertainty is ±0.005 g.
The scale shown below would display a mass of 20.00 g
Answer:
[H₂SO₄] = 6.07 M
Explanation:
Analyse the data given
8.01 m → 8.01 moles of solute in 1kg of solvent.
1.354 g/mL → Solution density
We convert the moles of solute to mass → 8.01 mol . 98g /1mol = 785.4 g
Mass of solvent = 1kg = 1000 g
Mass of solution = 1000g + 785.4 g = 1785.4 g
We apply density to determine the volume of solution
Density = Mass / volume → Volume = mass / density
1785.4 g / 1.354 g/mL = 1318.6 mL
We need this volume in L, in order to reach molarity:
1318.6 mL . 1L / 1000mL = 1.3186 L ≅ 1.32L
Molarity (mol/L) → 8.01 mol / 1.32L = 6.07M