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padilas [110]
3 years ago
6

How do i solve this? and please show out the working

Chemistry
2 answers:
irina1246 [14]3 years ago
4 0
What does it say maybe I can help u
Mrrafil [7]3 years ago
3 0
I don't know what does it means after “calculate the”
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What is an ionic compound
Gwar [14]
Ionic compound is formed by negative and positive ions, these ions form ionic bond
6 0
3 years ago
The ka of hypochlorous acid (hclo) is 3.0 ⋅ 10−8 at 25.0 °c. calculate the ph of a 0.0375m hypochlorous acid solution.
Scrat [10]
We can set up an ICE table for the reaction:                      
                      HClO          H+     ClO-
Initial              0.0375       0        0
Change         -x               +x      +x
Equilibrium    0.0375-x     x        x

We calculate [H+] from Ka:     
     Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)

Approximating that x is negligible compared to 0.0375 simplifies the equation to         
     3.0x10^-8 = (x)(x)/0.0375     
     3.0x10^-8 = x2/0.0375     
     x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9     
     x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:     
     pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
6 0
3 years ago
In which pair do both atoms have one electron only in an s orbital in their ground states?
Scrat [10]
I can help you only with what is electron in ground state. The electron which is not excited or over which no external force is acting that could possibly make it jump to the higher state,is said to be in its ground state.In excited state,the electron has more energy than it normally has. For convenience,suppose you are an electron,you are in Normal mood,but when your little brother snatches the Tv remote from your hand,you become angry.Your normal state is your ground state and your angry state is your excited state.
5 0
2 years ago
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150
mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after  5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ <em>C</em> NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ <em>C</em> CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ <em>C</em> CH3COOH = 0.0375 M

⇒ <em>C</em> NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ <em>C </em>NaOH = 0.075 M

⇒ <em>C</em> CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ <em>C </em>NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ <em>C </em>NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0
3 years ago
Is dimethyl amine and triethylamine more soluble in water?
kompoz [17]
No, dimethyl amine and triethyl amine are not more soluble in water. Both are soluble in water but not more soluble in water. These are also soluble in alcohol.
Dimethyl amine is a secondary amine while triethyl amine is a tertiary amine. Generally, the small amines are more soluble in water. As the hydrocarbon chain goes on increasing, the solubility goes on decreasing.
4 0
3 years ago
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