Answer : The voltage applied by the batteries is, 6.0 V
Solution : Given,
Resistance of flashlight = 2.4 ohm
Current in the circuit = 2.5 Ampere
Formula used :
where,
V = applied voltage
I = current in the circuit
R = resistance of light
Now put all the given values in the above formula, we get
Therefore, the voltage applied by the batteries is, 6.0 V
Answer:
Concentration of chloride ions = 0.584M
Explanation:
The step by step calculations is shown as attached below.
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
