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Natasha2012 [34]
3 years ago
10

A 25.00 mL aliquot of concentrated hydrochloric acid (11.7M) is added to 175.00 mL of 3.25M hydrochloric acid. Determine the num

ber of moles of hydrochloric acid from 175.00 mL of 3.25 M hydrochloric acid solution.
Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
5 0

Answer:

The number of moles of hydrochloric acid are 0.861

Explanation:

In first solution, the [HCl] is 11.7 M, which it means that 11.7 moles are present in 1 liter.

So we took 25 mL and we have to know how many moles, do we have now.

1000 mL ____ 11.7 moles

25 mL _____ (25 . 11.7)/1000 = 0.2925 moles

This are the moles, we add to the solution where the [HCl] is 3.25 M

In 1000 mL __ we have __ 3.25moles

175 mL ____ we have __ (175 . 3.25)/1000 = 0.56875 moles

Total moles: 0.2925 + 0.56875 = 0.861 moles

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If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Pure Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Br and Br,

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

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For N and O,

                    E.N of Oxygen      =   3.44

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                    E.N of Hydrogen       =   2.20

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For K and O,

                    E.N of Oxygen          =   3.44

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6 0
4 years ago
How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

<u>Step 2: Identify Conversions</u>

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

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Answer:

See Explanation

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One thing that we must have in mind is that it takes much more energy to remove an electron from an inner filled shell than it takes to remove an electron from an outermost incompletely filled shell.

Now let us consider the case of magnesium which has two outermost electrons. Between IE2 and IE3 we have now moved to an inner filled shell(IE3 refers to removal of electrons from the inner second shell) and a lot of energy is required to remove an electron from this inner filled shell, hence the jump.

For aluminium having three outermost electrons, there is a jump between IE3 and IE4 because IE4 deals with electron removal from a second inner filled shell and a lot of energy is involved in the process hence the jump.

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8 0
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Negatively charged particles in the outermost<br> energy level of the electron cloud.
balandron [24]

Answer:

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Explanation:

6 0
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