Answer is: sodium (Na) and iodine (I₂).
<span>
First ionic bonds in this salt are separeted
because of heat:
</span>NaI(l) → Na⁺(l) + I⁻(l).
Reaction of reduction
at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.
2Na⁺(l) + 2e⁻ → 2Na(l).
Reaction of oxidation
at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.
The anode is positive
and the cathode is negative.
Answer : The correct option is, (2) Cr (Chromium)
Explanation :
The reactivity series of metal are arranged of the reactivity from the highest to the lowest. Reactivity series is used to determine the products of the single displacement reactions. In the single displacement reaction, the most reactive metal displaces the least reactive metal.
From the given reactivity series we conclude that there are two metal (Mg and Cr) are more reactive metal than the Ni and there are two metal (Pb and Cr) are less reactive metal than the Zn. So, the Cr (Chromium) is the metal which is more active than Ni and less active than Zn.
Hence, the correct option is, (2) Cr
Answer is C.
Converting temperatures to degrees K: 33 degrees C = 306 K and -55 = 218 degrees K.
By the ideal gas law:-
760 * 0.50 / 306 = P * 0.10 / 218
P = 760 * 0.50 * 218 / 306 * 0.10
= 2700 mm Hg answer
Answer:
A saturated solution can become supersaturated when it is cooled. The solubility of solid solutes in liquid solvents increases as the solvent is warmed up. For example, you can dissolve more sugar in warm water as opposed to cold water.
Answer:
1.09 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>
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The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:
100 g × (1 mol/255.43 g) = 0.391 mol
0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:
0.391 mol × (1 L/0.360 mol) = 1.09 L
