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3 years ago

Need help on some homework questions I’m stuck on about molarity and dilutions:

Chemistry

1 answer:

Flura [38]3 years ago

6 0

question 1

The moles of C12H22O11 that are in 7.5 L of a 5.8 M C12H22O11 solution is 43.5 moles

calculation

moles = molarity x volume

volume = 7.5 L

molarity = 5.8 M or 5.8 mol/l

moles is therefore = 7.5 l x 5.8 mol/l = 43.5 moles

question 2

% percent =( mas of the solute/ mass of the solution) x 100

mass of solute= ?

mass of the solution = 375 g

% mass = 85 % = 85/100

let the mass of solute be represented by Y

therefore y/375 = 85/100

y=(375 x 85)/100 = 318.75 g

moles = mass/molar mass

= 318.75 g/ 42.4= 7.52 moles

Question 3

The grams of copper metal is 857.25 g

calculate the moles HCl

= molarity x volume = 4.50 L x 6 .00 mol/l= 27 moles

By use of mole ratio of Cu: HCl which is 1:2 the moles of Cu = 27/2 =13.5 moles

mass = moles x molar mass

= 13.5 moles x 63.5 g/mol = 857.25 g

question 4

Complete description of how to make 3.5 L of 2.0 M NaCl solution from a stock of 6.0 M NaCl is that measure 1.17 L of 6.0 M NaCl, Dissolve it to 3.5 L to make 2.0 M NaCl

calculation

M1V1 = M2V2

m1= 2.0 M

V1=3.5 L

M2= 6.0 M

V2=?

v2 = M1V1/M2

= (3.5 L x 2.0 M)/ 6.0 M= 1.17 L

explanation : measure 1.17 L of 6.0M NaCl and then dissolve it to 3.5 L to make 2.0 M NaCl

question 5

The number of grams of stock solution that is in 90.0 % is HCl by mass that would be needed to make 175 g is 77.78 grams

calculation

This is calculated using M1C1= M2C2 formula

M1 = ?g

C1 =90.0%

M1= 175 g

C1= 40.0 %

make M1 the subject of the formula

M1= M2C2/C1

M1 is therefore =( 175 g x 40) / 90 = 77.78 g

Question 6

The grams of silver metal is 127 g

% mass = mass of solute/ mass of solution

% mass = 80% = 80/100

mass of solute (AgNO3)= ?

mass of solution = 250 g

let the mass of solute be represented by Y

therefore Y/ 250 = 80/100

Y= (250 x 80) /100 =200 g of AgNO3

moles = mass/molar mass

moles of AgNO3 = 200 g/ 169.87 g/mol =1.178 moles

The mole ratio of AgNO3: Ag is 2:2=1:1 therefore the moles of Ag= 1.178 moles

mass= moles x molar mass

= 1.178 moles x 107.87 g/mol =127.07 g

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where $X_A$ & $X_B$ are the Pauling's electronegativities.

The table attached has the values of electronegativities, by taking the values of Al and Mn from there,

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Now, there are 3 types of inter atomic bonding

1) Ionic Bonding: It refers to the chemical bond in which there is complete transfer of valence electrons between atoms

2) Covalent Bonding: It refers to the chemical bond involving the sharing of electron pairs between 2 atoms.

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In $Al_6Mn$ compound, the % ionic character is minimal that is $\sim$ 0.20% and there are two metal ions present, therefore this compound will have metallic bonding.

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