Answer:
Damian here! (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧
The newly hatched larva is in its first instar, a developmental stage that occurs between molts. It feeds until it grows too big for its cuticle, or soft shell, and then it molts. After molting, the larva is in the second instar. Ladybug larvae usually molt through four instars, or larval stages, before preparing to pupate.
Explanation:
hope this helps? :))
Answer:
15.4%
Explanation:
If Ka = 0.54 mM = 1.51x10⁻⁵
Then;
C₄H₈O₂ --------> C₄H₇O₂⁻ + H⁺
I 0.54x10⁻³ 0 0
E 0.54x10⁻³(1-x) 0.54x10⁻³x 0.54x10⁻³x
Recall that x is the percentage degree of dissociation
From the ICE table;
Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]
1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)
1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x
1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2
1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2
Hence;
0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0
x^2 + 0.028x - 0.028 = 0
Solving the quadratic equation here;
x = 0.154 or −0.182
Ignoring the negative result, x = 0.154
Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%
Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
Regards!
Answer:
FeCl2 + H. 2
Explanation:
Iron react with hydrogen chloride to produce iron(II) chloride and hydrogen.
It could be used as a buffer, since it is not a conductor. Hope this helps!!!
