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3 years ago

What is effective nuclear charge?

Chemistry

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer:

The effective nuclear charge (often symbolized as Zeff or Z*) is the net positive charge experienced by an electron in a multi-electron atom. The term “effective” is used because the shielding effect of negatively charged electrons prevents higher orbital electrons from experiencing the full nuclear charge.

Explanation:

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A gas sample with a mass of 12.8 g exerts a pressure of 1.2 atm at 15°c and a volume of 3.94 l. what is the molar mass of the ga

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We can use the ideal gas law equation to find the number of moles in the gas
PV = nRTwhere P - pressure - 1.2 atm x 101 325 Pa/atm = 121 590 Pa
V - volume - 3.94 x 10⁻³ m³
n - number of moles
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substituting the values in the equation
121 590 Pa x 3.94 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
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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.

lisov135 [29]

Answer:

1. Molarity of MgSO₄ = 0.6 M

2. Molarity of AgNO₃ = 0.06 M

3. volume of acetic acid = 250 mL

4. Molarity of NaCl= 2.3 M

5. Mass of C₁₂H₂₂O₁₁ = 4.1 g

6. Mass of C₁₀H₈ (naphthalene) = 77 g

7. mass of ethanol = 11 g

8. Mass of DDT = 0.011 mg

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.

Mass of C₁₂H₂₂O₁₁ = 4.1 g

____________________

The remaing portion is in attachment

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4 years ago