The independent variable is the variable being changed. In this case, the independent variable is the calculators. The dependent variable is essentially what you are looking for that <u>depends</u> on the independent variable. In this case it would be time. The constant variable or controlled variable are something that doesn't change and would skew the results. One may be the exact same problem for both groups. Try to come up with two more.
Answer:
Due to presence of a triple bond between the two N−atoms, the bond dissociation enthalpy (941.4 kJ mol
−1
) is very high. Hence, N
2
is the least reactive.
Answer:
C) 1 x 10-10 M
Explanation:
To solve this question we must use the equation:
Kw = [H+] [OH-]
<em>Where Kw is the equilibrium dissociation of water = 1x10-14</em>
<em>[H+] is the molar concentration of hydronium ion = 1x10-4M</em>
<em>[OH-] is the molar concentration of hydroxyl ion</em>
<em />
Replacing:
1x10-14= 1x10-4 [OH-]
<em>[OH-] = 1x10-14 / 1x10-4M</em>
<em>[OH-] = 1x10-10 M</em>
Right option is:
<h3>C) 1 x 10-10 M
</h3>
Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry)

= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
The answer is a strike-slip. More specifically a right-lateral strike-slip.