Answer:
1.06 V
Explanation:
The standard reduction potentials are:
Ag^+/Ag E° = 0.7996 V
Ni^2+/Ni E° = -0.257 V
The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are
Ni → Ni^2+ + 2e- E° = 0.257 V
<u>2Ag^+ 2e- → 2Ag </u> <u>E° = 0.7996 V
</u>
Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V
To three significant figures, the standard potential for the cell is 1.06 V
.
Explanation:
Hardness test — Scratch the rock with a fingernail, a copper penny, a glass plate or nail, and a ceramic plate. Check your Guide to assign it a rating on the Mohs Scale of Hardness.
Color streak test — Test for the “color streak” of the minerals by rubbing the rock across the ceramic plate in the Mineral Test Kit, or across smooth
cement. Look up which colors indicate which minerals are present.
Magnetism test — Hold the magnet in the Mineral Test Kit near your rock. If there is a magnetic pull, it has a metal mineral in it.
Acidity test — Put vinegar in the bottle included in the Mineral Test Kit. Squeeze out a few drops on the rock. If it fizzes, it contains carbonate.
A quick and easy way to find out whether your diamond is real or fake: try fogging it up with your breath. If it clears up after one or two seconds, then your diamond is real, but if it stays fogged for three to four seconds chances are that you're looking at a fake.
Heat would be required : 1,670 J
<h3>Further explanation</h3>
Given
mass of H₂O=5 g
Required
Heat to melt
Solution
The heat to change the phase can be formulated :
Q = m.Lf (melting/freezing)
Lf=latent heat of fusion
The heat of fusion for water at 0 °C : 334 J/g
Input given values in formula :
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
Answer:
Using cobalt glass could be helpful to identify elements that weakly emit blue and/or violet.
Explanation:
