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irinina [24]
3 years ago
13

Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC plate

and developed in a hexanes-ethyl acetate solvent mixture. Predict the relative Rf values for the three compounds in the sample (biphenyl, benzoic acid, and benzyl alcohol).
Chemistry
1 answer:
Citrus2011 [14]3 years ago
4 0

Answer:Consider a sample that is a mixture composed of biphenyl, benzoic acid, and benzyl alcohol. The sample is spotted on a TLC ... Predict the relative Rf values for the three components in the sample. Hint: See Table 19.3. ... The sample is spotted on a TLC plate and developed in solvent mixture. We are going to predict the ...

Explanation:MORE POWER

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4. A salvage operator recovered coins believed to be gold. A sample weighed 385.000g and has a volume of 20.0mL. Were the coins
aleksley [76]

Answer:

Gold

Explanation:

We are given that

Mass of sample ,m=385 g

Volume ,V=20mL

We have to find the coin is gold or yellow brass.

We know that

Density=\frac{Mass}{volume}

Using the formula

Density=\frac{385}{20}g/mL

Density=19.25g/mL

Density\approx 19.3g/mL

Density of coin=19.3g/mL

Density of gold=19.3g/mL

Hence, the coin is gold.

5 0
3 years ago
A substance joined by this type of bond has a low melting point
cestrela7 [59]
Covalent compounds have bonds where electrons are shared between atoms. Due to the sharing of electrons, they exhibit characteristic physical properties that include lower melting points and electrical conductivity compared to ionic compounds.
5 0
2 years ago
Read 2 more answers
Substance formed by the chemical combination of elements is called
nikitadnepr [17]

Answer:

chemical equation

Explanation:

3 0
3 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
ExtremeBDS [4]

Answer:

1597.959 g  

Explanation:

Given Data:

Amount of Cr₂(SO₄)₃ = 450 g

Amount of potassium phosphate K₃PO₄ = in Excess

grams of potassium sulfate K₂SO₄= ?

Solution

The Reaction will be

                 Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

Information that we have from reaction

                Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

                    1 mol          2 mol                   3 mol

we come to know from the above reaction that

1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

We also know that

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

if we represent mole in grams then

      Cr₂(SO₄)₃             +       2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

       1 mol (147 g/mol)         2 mol  (212 g/mol)      3 mol  (174g/mol)

So, Now we have the following details

          Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

              147 g         424 g                           522 g

So,

we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce  522 g of K₂SO₄

So now we calculate that how many grams of potassium sulfate will be produced

Apply unity formula

              147 g of  Cr₂(SO₄)₃  ≅ 522 g of K₂SO₄

              450 g of  Cr₂(SO₄)₃  ≅ ? g of K₂SO₄

by doing cross multiplication

g of K₂SO₄ =522 g x 450 g / 147 g

g of K₂SO₄ =  1597.959 g

So the write answer is  1597.959 g  

***Note: By calculation it is obvious that the correct answer is  1597.959 g  

8 0
3 years ago
The standard emf for the cell using the overall cell reaction below is +2.20 V:
ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: 2Al-6e^{-}\rightarrow 2Al^{3+}

Reduction: 3I_{2}+6e^{-}\rightarrow 6I^{-}

---------------------------------------------------------------------------------

Overall: 2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}

Nernst equation for this cell reaction at 25^{0}\textrm{C}-

E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}

where n is number of electrons exchanged during cell reaction, E_{cell}^{0} is standard cell emf , E_{cell} is cell emf , [Al^{3+}] is concentration of Al^{3+} and [Cl^{-}] is concentration of Cl^{-}

Plug in all the given values in the above equation -

E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V

So, E_{cell}=2.32V

3 0
2 years ago
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