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jarptica [38.1K]
2 years ago
14

Bonding Story Mini Project

Chemistry
1 answer:
ioda2 years ago
4 0
There needs to be the story so that i can work with the 1st question
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Need help! Don’t understand
shtirl [24]
Bro honestly I don’t understand either
8 0
3 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.
Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = pK_{a} + log \frac{Base}{Acid}  

          4.245 = 3.75 + log \frac{Base}{Acid}

      log \frac{Base}{Acid} = 0.5

    \frac{Base}{Acid} = 3.162

Now,

        \frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

                 

8 0
3 years ago
A form of electricity which can attract things
qwelly [4]

 

A form of electricity which can attract things is static electricity. Static electricity is the result of an imbalance between negative and positive charges in an object. These charges built up o the surface of an object until they find a way to e released or discharged causing the attraction of things.

8 0
3 years ago
In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o
Zigmanuir [339]

Answer:

Stock  solution =  1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

<u>Step 1:</u> Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

<u>Step 2</u>: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

<u>Step 3:</u> Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

<u>Step 4</u>: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

<u>Step 5:</u> Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

<u>Step 6</u>: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0
3 years ago
When bonds are (broken/formed) there is a positive energy change.
krek1111 [17]

Answer: Hello i am confused are you asking a question?

Explanation:

3 0
2 years ago
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