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mixer [17]
3 years ago
7

Which 0.1 M aqueous solution is the best conductor of electricity /12561947/e9552934?utm_source=registration

Chemistry
1 answer:
Andrej [43]3 years ago
3 0
Above question is incomplete. Complete question is:
<span>Which 0.1-molar aqueous solution is the best conductor of electricity? (1) H2S (2) H2SO4 (3)HF (4)H3PO4?
...........................................................................................................................

Answer: Correct answer is option 4, i.e. H3PO4.

Reason:
Each molecule of H3PO4 on dissociation will generate 4 ions (three H+ ions and one PO4 3-ion). On other hand, H2SO4 and HF will generate three and two ions respectively. H2S is a covalent compound, and will not generate any ions in solution. Due to largest number of ions, 0.1 M H3PO4 solution will have maximum conductivity. Hence, it will be the best conductor of electricity. 

</span>
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Magnets can be used in space. ... One class of magnets, called electromagnets, does need electricity to work.

Explanation:

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2. What are the different areas of an electron cloud called?
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Hello &lt;3 I just need some help with periodic table of elements.
garik1379 [7]

Answer:

11

Explanation:

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8 0
2 years ago
Read 2 more answers
Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5
FinnZ [79.3K]
The answer for this issue is: 
The chemical equation is: HBz + H2O <- - > H3O+ + Bz- 
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz] 
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x. 
Accept that x is little contrasted with 0.5 M. At that point, 
Ka = 6.4X10^-5 = x^2/0.5 
x = [H3O+] = 5.6X10^-3 M 
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make) 
3 0
3 years ago
A solution is prepared by mixing 50.0 mL toluene (C6H5CH3 d=0.867 g/mL) with 125 mL Benzene (C6H6 d=0.874 g/mL). Assuming that t
Tatiana [17]

Answer:

mass % = 28.4%

mole fraction = 0.252

molality = 3.08 molal

molarity = 2.69 M

Explanation:

Step 1: Data given

Volume of toluene = 50.0 mL = 0.05 L

Density toluene = 0.867 g/mL

Molar mass toluene = 92.14 g/mol

Volume of benzene = 125 mL = 0.125 L

Density benzene = 0.874 g/mL

Molar mass benzene = 78.11 g/mol

Step 2: Calculate masses

Mass = density * volume

Mass toluene = 50.0 mL * 0.867 g/mL = 43.35 g

Mass benzene = 125 mL * 0.874 g/mL = 109.25 g

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles toluene = 43.35 grams /92.14 g/mol = 0.470 5 moles

Moles benzene = 109.25 grams / 78.11 g/mol = 1.399 moles

Step 4: Calculate molarity of toluene

Molarity = moles / volume

Molarity toluene = 0.4705 moles / 0.175 L = 2.69 M

Step 5: Calculate mass % of toluene

Mass % = (43.35 grams / (43.35 + 109.25) )*100 % = 28.4 %

Step 6: Calculate mole fraction of toluene

Mole fraction toluene = Moles toluene / total number of moles

Mole fraction toluene = 0.4705 / (0.4705 + 1.399) = 0.252

Step 7: Molality of toluene

Molality = number of moles / mass

Molality of toluene = 0.4705 moles / (0.04335 + 0.10925)

Molality of toluene = 3.08 molal

4 0
3 years ago
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