Answer:
1.
was the
value calculated by the student.
2.
was the
of ethylamine value calculated by the student.
Explanation:
1.
The
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = ![\frac{2.00 g}{180 g/mol}=0.01111 mol](https://tex.z-dn.net/?f=%5Cfrac%7B2.00%20g%7D%7B180%20g%2Fmol%7D%3D0.01111%20mol)
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = ![\frac{0.01111 mol}{0.600 L}=0.0185 M](https://tex.z-dn.net/?f=%5Cfrac%7B0.01111%20mol%7D%7B0.600%20L%7D%3D0.0185%20M)
![HAs\rightleftharpoons As^-+H^+](https://tex.z-dn.net/?f=HAs%5Crightleftharpoons%20As%5E-%2BH%5E%2B)
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:
![K_a=\frac{x\times x }{(c-x)}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7Bx%5Ctimes%20x%20%7D%7B%28c-x%29%7D)
![=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.00240%20M%5Ctimes%200.00240%20M%7D%7B%280.0185-0.00240%20%29%7D)
![K_a=3.57\times 10^{-4}](https://tex.z-dn.net/?f=K_a%3D3.57%5Ctimes%2010%5E%7B-4%7D)
was the
value calculated by the student.
2.
The
value of ethylamine = 11.87
![pH+pOH=14](https://tex.z-dn.net/?f=pH%2BpOH%3D14)
![pOH=14-11.87=2.13](https://tex.z-dn.net/?f=pOH%3D14-11.87%3D2.13)
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M
![C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-](https://tex.z-dn.net/?f=C_2H_5NH_2%2BH_2O%5Crightleftharpoons%20C_2H_5NH_3%5E%7B%2B%7D%2BOH%5E-)
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:
![K_b=\frac{x\times x}{(c-x)}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7B%28c-x%29%7D)
![=\frac{0.00741\times 0.00741}{(0.100-0.00741)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0.00741%5Ctimes%200.00741%7D%7B%280.100-0.00741%29%7D)
![K_b=5.93\times 10^{-4}](https://tex.z-dn.net/?f=K_b%3D5.93%5Ctimes%2010%5E%7B-4%7D)
was the
of ethylamine value calculated by the student.
Answer:
A sample of helium gas has a volume of 620mL at a temperature of 500 K. If we ... to 100 K while keeping the pressure constant, what will the new volume be?
Explanation:
Answer:
Decrease
Explanation:
Since the speed in which the gas molecules are faster as they are heated, they fly around in the container and logically, it is harder to insert a moving object into water than something more stationary or slower.
Answer:
b. Conducts electricity when dissolved in water
Explanation:
Iron(II) chloride, is the chemical compound with formula FeCl2.
It is a solid with a high melting point of about 677 degree Celsius or 950 K when in anhydrous form but have lower melting point in hydrated form.
The compound is often off-white. FeCl2 crystallizes from water as the greenish tetrahydrate, which is the form that is most commonly encountered in the laboratory.
There is also a dihydrate. The compound is highly soluble in water, giving pale green solutions.
The compound : C₄₀H₄₄N₄O
<h3>Further explanation</h3>
The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components
The principle of determining empirical formula
• Determine the mass ratio of the constituent elements of the compound.
• Determine the mole ratio by dividing the percentage by the atomic mass
The mol ratio of composition : C : H : N : O
![\tt \dfrac{80.66}{12}\div \dfrac{7.39}{1}\div \dfrac{9.39}{14}\div \dfrac{2.68}{16}\\\\6.722\div 7.39\div 0.671\div 0.1675\rightarrow divide~by~smallest(0.1675)\\\\40\div 44\div 4\div 1](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B80.66%7D%7B12%7D%5Cdiv%20%5Cdfrac%7B7.39%7D%7B1%7D%5Cdiv%20%5Cdfrac%7B9.39%7D%7B14%7D%5Cdiv%20%5Cdfrac%7B2.68%7D%7B16%7D%5C%5C%5C%5C6.722%5Cdiv%207.39%5Cdiv%200.671%5Cdiv%200.1675%5Crightarrow%20divide~by~smallest%280.1675%29%5C%5C%5C%5C40%5Cdiv%2044%5Cdiv%204%5Cdiv%201)