Answer: A
Aloha is the one that loses a helium
Answer:They are small because they don't need larger bodies to enable them attach to the females. They don't have fully formed gut because when they reach adulthood their digestive system stops functioning.
Explanation:they then find a female angler fish, attach by bitting into her flesh and fusing to her body. So whatever she eats they eat too.
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>
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
![\frac{[H^+][HC2O4^-]}{[H2C2O4]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BHC2O4%5E-%5D%7D%7B%5BH2C2O4%5D%7D%20)
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
Answer:
2.445 g
Explanation:
Step 1: Given and required data
- Energy in the form of heat required to boil the water (Q): 5525 J
- Latent heat of vaporization of water (∆H°vap): 2260 J/g
Step 2: Calculate the mass of water
We will use the following expression.
Q = ∆H°vap × m
m = Q / ∆H°vap
m = 5525 J / (2260 J/g)
m = 2.445 g