Answer:

Explanation:
<em>Ferrous Sulphate</em>
<em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>
<em>, Sulphur Dioxide </em>
<em> as well as Ferric Oxide </em>
<em>.</em>
<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>
<em>Hence,</em>

<em>Now,</em>
<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>
<em>1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>
.
<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>
<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>
<em> </em>
<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
Answer: Concentration of
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of
= 0.729 M
The given balanced equilibrium reaction is,

Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :



concentration of
in the equilibrium mixture = 
Thus concentration of
in the equilibrium mixture is 0.31 M
The answer is number 2. That releases massive amounts of radiation and by the way, that is how atomic bombs are made to detonate.
Answer:
In cellular respiration, cells use <u>oxygen</u> to release energy stored in <u>glucose</u>.
If my answer helped, please mark me as the brainliest!!
Thank You!!