Answer:
- Empirical:
- Molecular:
Explanation:
Hello,
In this case, based on the information regarding the combustion, the moles of carbon turn out:
Moreover, the moles of hydrogen:
Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:
Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:
In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:
Which is hexane.
Best regards.
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: Heating a crucible to remove water from a hydrate.
Explanation:
The options are:
a. Heating a solvent to help a solute dissolve.
b. Heating an isolated solid to dry it.
c. Heating water to boiling for a water bath.
d. Heating a crucible to remove water from a hydrate.
The procedure that can be performed on a hot plate are:
a. Heating a solvent to help a solute dissolve.
b. Heating an isolated solid to dry it.
c. Heating water to boiling for a water bath.
It should be noted that the hot plate cannot be used for heating of crucible in order to remove water from a hydrate. It is not advisable for someone to heat any silica or ceramic objects on a hot plate.
Therefore, heating a crucible to remove water from a hydrate is the correct option.
Answer:
pH = 9.6
Explanation:
According to Brönsted-Lowry theory, NH₃ is a base and NH₄⁺ its conjugate acid. When they are together in a solution, the form a buffer, which is used to resist abrupt changes in pH when an acid or a base is added. pOH fro a buffer can be found using Henderson-Hasselbalch equation.
Since NH₄Cl is a strong electrolyte, [NH₄Cl] = [NH₄⁺]
![pOH = pKb + log\frac{[NH_{4}^{+} ]}{[NH_{3}]} =4.7+log\frac{0.035M}{0.070M} =4.4](https://tex.z-dn.net/?f=pOH%20%3D%20pKb%20%2B%20log%5Cfrac%7B%5BNH_%7B4%7D%5E%7B%2B%7D%20%5D%7D%7B%5BNH_%7B3%7D%5D%7D%20%3D4.7%2Blog%5Cfrac%7B0.035M%7D%7B0.070M%7D%20%3D4.4)
Now, we can find pH using the following expression:
pH + pOH = 14
pH = 14 - pOH = 14 - 4.4 = 9.6
