The balanced reaction that describes the reaction between sulfuric and sodium hydroxide to produce sodium sulfate and water is expressed <span>H2SO4 + 2NaOH → 2H2O + Na2SO4. When 65.5 grams of sodium hydroxide reacts with excess sulfur, we need first to convert the mass to moles and multiply by stoich ratio 2/2 or 1. Hence the moles is multiplied to the molar mass of water. The final answer is 29.475 grams water.</span>
Answer:
230
Th
90
Explanation:
The left side shows the element uranium with mass number 234 (superscript to the left of ths symbol) and atomic mass 92 (subscript to the left of the symbol).
So, you have to find the mass number and atomic numbers missing on the right side which added to those shown for the helium atom reach 234 and 92.
This the the mathematics:
1) Mass number of U - mass number of He = 234 - 4 = 230 ⇒ the mass number of the missing element is 230.
2) atomic number of U - atomic number of He = 92 - 2 = 90 ⇒ the atomic number of the missing element is 90.
Now, you use the atomic mass to identify the element. You do it by simply searching in a periodic table. There you find that the atomic number 90 corresponds to thorium, whose symbol is Th.
Now you have that the product that completes the right side of the decay equation is:
230
Th
90
A, B, and C because they all equal 5.6 cm and it’s not close to 6.

where, E^{o} (Ag+/Ag) = std. reduction potential of Ag+ = 0.7994 v
and Sn2+/Sn = std. reduction potential of Sn2+ = -0.14 v
Thus, E^{o}cell = 0.7994v - (-0.14v) = 0.9394 v
Now, ΔG^{o} = -nF

,
where, n = number of electrons = 2
F = Faraday's constant = 96500 C
∴ΔG^{o} = 2 X 96500 X 0.9394 = -1.18 X

Now, using Nernst's Equation we have,
![[tex]E_{cell} = 0.9394 - \frac{2.303X298}{2X96500}log \frac{0.0115}{ 3.5^{2} }](https://tex.z-dn.net/?f=%20%5Btex%5DE_%7Bcell%7D%20%3D%200.9394%20-%20%5Cfrac%7B2.303X298%7D%7B2X96500%7Dlog%20%5Cfrac%7B0.0115%7D%7B%203.5%5E%7B2%7D%20%7D%20)
E_{cell} = 0.9765 v
Finally, ΔG = -nFE = -2 X 96500 X 0.9765 = -1.88 X
Water and salt are the products of a neutralization reaction.