Answer:
CuSO4 cell will have the greatest amount of deposit among all three. The deposit will occur at the cathode
Explanation:
The valence of the elements in this case is as follows -
Cu - 2e-
Sn - 4e-
Cr - 3e-
CuSO4 cell will have the greatest amount of deposit among all three
The atoms of copper metal will deposit at the cathode. At the cathode, the least number of moles of electrons needed .
Hence, more amount of copper can be extracted out by the electrolyte
Physical because you can see it evaporate have a great día
Answer:
The answer is "2%"
Explanation:
Equation:


Formula:
![Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BNO_2%5E%7B-%7D%5D%7D%7B%5BHNO_2%5D%7D)
Let
at equilibrium

therefore,
![[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%202.0%5Ctimes%2010%5E%7B-2%7D%20%5C%20M%20%3D%200.02%20%5C%20M)
Calculating the % ionization:
![= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%5BH%5E%7B%2B%7D%5D%7D%7B%5BHNO_2%5D%29%7D%20%5Ctimes%20100%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.02%7D%7B1%7D%5Ctimes%20100%20%5C%5C%5C%5C%3D%202%5C%25%5C%5C%5C%5C)
Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, 
Explanation:
The reaction equation for given reaction is as follows.

Here, 1 mole of
reacts with 3 moles of
.
As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide
is 152 g/mol.
Number of moles is the mass of substance divided by its molar mass. So, moles of
is calculated as follows.

Now, moles of
.given by 0.5 mol of
is calculated as follows.

As molar mass of
is 2.016 g/mol. Therefore, mass of
is calculated as follows.

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide,
.
First, they may produce direct and relatively short-lived effects, such as stimulation of the isolated uterus or relaxation of the isolated tracheal chain preparation.
Secondly, in doses too low to produce a direct effect, they may produce a long-term potentiation of the effects of other stimulants.