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tino4ka555 [31]
3 years ago
14

In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical

formula of the compound?
Chemistry
1 answer:
Umnica [9.8K]3 years ago
3 0

Grams of Phosphorus = 4.14 grams 
Grams of white compound = 27.8 grams 
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
 Calculating moles which would be grams / molar mass
 Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
 Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
 Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
 Calculating the ratios by dividing with the small entity
 P = 0.1337 moles / 0.1337 moles = 1
 Cl = 0.6674 moles / 0.1337 moles = 5 
So the empirical formula would be PCl5
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Each answer below describes the connectivity around a single atom.
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Answer:

d. One single bond and two double bonds.

Explanation:

The octate rule is a chemical rule in which the atoms prefer to have eight electrons in the valence shell. Where a single bond provide two electrons and a double bond provide 4 electrons. Thus:

a. Two double bonds . Two double bonds provide 8 electrons. Octate rule <em>is not </em>violated

b. Three single bonds and one pair of electrons . Three single bonds provide 6 electrons and one pair of electrons provide two electrons. Thus, you have eight electrons and octate rule <em>is not</em> violated

c. Two single bonds and one double bond . Two single bonds provide four electrons and one double bond 4. Thus, you have eight electrons and octate rule <em>is not </em>violated.

d. One single bond and two double bonds. One single bond provides two electrons and two double bonds 8. Thus, you have 10 electrons and <em>octate rule is violated.</em>

e. Four single bonds. Four single bonds provide 8 electrons. Octate rule<em> is not </em>violated.

I hope it helps!

4 0
3 years ago
Part C<br> Number of molecules in 8.437x10-2 mol C6H6
N76 [4]

Answer:

There are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

Explanation:

In this problem, we need to find the number of molecules in 8.437\times 10^{-2} mol of C_6H_6.

The molar mass of C_6H_6 is 6\times 12+1\times 6=78\ g/mol

No of moles = mass/molar mass

We can find mass from above formula.

m=n\times M\\\\m=8.437\times 10^{-2}\ mol\times 78\ g/mol\\\\m=6.58\ g

Also,

No of moles = no of molecules/Avogadro number

N=n\times N_A\\\\N=8.437\times 10^{-2}\times 6.023\times 10^{23}\\\\N=5.08\times 10^{22}\ \text{molecules}

Hence, there are 5.08\times 10^{22}\ \text{molecules of}\ C_6H_6  

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Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

The percent yield

Solution

mol AgNO₃(MW=169,87 g/mol) :

= mass : MW

= 127 : 169.87

= 0.748

mol Ag from equation :

= 2/2 x mol AgNO₃

= 2/2 x 0.748

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<em>%yield = 88.5%</em>

7 0
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