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BlackZzzverrR [31]
3 years ago
12

Match each decimal number to the correct scientific notation. 30,700 0.000307 0.00000307 3,070,000 3.07 × 10-6 arrowRight 3.07 ×

106 arrowRight 3.07 × 10-4 arrowRight 3.07 × 104 arrowRight
Chemistry
2 answers:
ss7ja [257]3 years ago
8 0

Answer:

3.07 × 10^4=30,700

3.07 × 10^-4=0.000307

3.07 × 10^-6=0.00000307

3.07 × 10^6=3,070,000

Explanation:

patriot [66]3 years ago
6 0

Answer:

  1. 30,700 = 3.07\times 10^4
  2. 0.000307 =3.07\times 10^{-4}
  3. 0.00000307 = 3.07\times 10^{-6}
  4. 3,070,000 = 3.07\times 10^6

Explanation:

Scientific notation: It is the representation of expressing the numbers which are too big or too small by representing them in decimal form with one digit before the decimal point times 10 raise to the power.

For example : 6,000,000 is written as 6.0\times 10^6

  1. 30,700 = \frac{10^4}{10^4}30,700=3.07\times 10^4
  2. 0.000307 =\frac{000307}{10^6}= 3.07\times 10^{-4}
  3. 0.00000307 = \frac{00000307}{10^8}= 3.07\times 10^{-6}
  4. 3,070,000 = \frac{10^6}{10^6 }\times 3,070,000=3.07\times 10^6
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1. Calculate the molarity when 403 g MgSO4 is dissolved to make a 5.25 L solution. Round to two significant digits.
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Answer:

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

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Explanation:

Ans 1.

Data given

mass of MgSO₄ = 403

Volume of solution = 5.25 L

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of MgSO₄ = 24 + 32 + 4(16)

Molar mass of MgSO₄ = 120 g/mol

Put values in equation 1

             no. of moles = 403 g / 120 g/mol

             no. of moles = 3.36 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 3.36 mol / 5.25 L

           M = 0.64

So, round the figure to two significant digits.

Molarity of MgSO₄ = 0.64 M

_____________________

Ans 2.

Data given

mass of AgNO₃ = 1.24 g

Volume of solution = 125 mL

convert mL to L

1000 mL = 1 L

125 mL = 125/1000 = 0.125

Solution:

First we will find mole of solute

                 no. of moles = mass in grams / Molar mass . . . . . .(1)

Molar mass of AgNO₃ = 108 + 14 + 3(16)

Molar mass of AgNO₃ = 170 g/mol

Put values in equation 1

             no. of moles = 1.24 g / 170 g/mol

             no. of moles = 0.0073 mol

Now to find molarity

Formula used

            Molarity = No. of moles of solute / solution in L . . . . . . (2)

Put Values in above equation

           M = 0.0073 mol / 0.125 L

           M = 0.06

So, round the figure to two significant digits.

Molarity of AgNO₃ = 0.06 M

______________________

Ans 3.

Data Given:

volume of acetic acid = 500 mL

% solution of acetic acid (M)= 50%

volume of acetic acid needed = ?

Solution:

formula used

    percent of solution = volume of solute/ volume of solution x 100

Put Values in above formula

                        50 % = volume of solute / 500 mL x 100

Rearrange the above equation

                   volume of solute =   50 x 500 mL /100

                   volume of solute =   250 mL

So, volume of acetic acid = 250 mL

______________________

Ans 4

Data Given:

Molarity of NaCl (M1) = 6 M

Volume of NaCl (V1) = 750 mL

convert mL to L

1000 ml = 1 L

750 ml = 750/1000 = 0.75 L

Volume of dilute solution (V2)= 2 L

Molarity of dilute solution (M2) = ?

Solution:

Dilution Formula will be used

                M1V1 = M2V2 . . . . . . (1)

Put values in equation 1

               6  M x 0.75 L = M2 x 2 L

Rearrange the above equation

               M2 = 6 M x 0.75 L / 2 L

               M2 = 2.25 M

So, round the figure to two significant digits.

Molarity of NaCl= 2.3 M

_________________________

Ans 5.

Data Given:

Molarity of C₁₂H₂₂O₁₁ = 0.16 M

Mass of C₁₂H₂₂O₁₁ (m) = ?

Volume of dilute solution = 75 mL

convert mL to L

1000 ml = 1 L

75 ml = 75/1000 = 0.075 L

Solution:

As we know

            Molarity = no.of moles/ liter of solution . . . . . . .(1)

we also know that

           no.of moles = mass in grams / molar mass . . . . . (2)

Combine both equation 1 and 2

            Molarity = (mass in grams / molar mass) / liter of solution . . . . (3)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 12(12) +22(1) +11(16)

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 144 +22 + 176

Molar mass of C₁₂H₂₂O₁₁ (Sucrose) = 342 g/mol

Put values in equation in equation 3

              0.16 M = (mass in grams / 342 g/mol) / 0.075 L

Rearrange the above equation

         mass in grams = 0.16 mol/L x 0.075 L x 342 g/mol

         mass in grams = 4.104 g

So, round the figure to two significant digits.

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