Water molecules have a dipole moment of 6.2 x 10-30 C-m because oxygen is more electronegative...

Water molecules have a dipole moment of 6.2 x 10-30 C-m because oxygen is more electronegative than hydrogen. If a water molecule sits at the origin and an electron sits at x= 10 nm, what is the magnitude of the force on the electron in fN? "f" is for femto b. In which direction is the force on the electron at x = 10 nm? c. What is the force on an electron that sits at y = 10 nm in fN? d. In which direction is the force on the electron at y = 10 nm?

Expert Solution

genius_generous answered 2 months ago

HCN has a pKa = 6.2 x 10-10. If a 50.0 mL of 0.100 M HCN...

HCN has a pKa = 6.2 x 10-10. If a 50.0 mL of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH a) after 8.00 mL base, b) at the halfway point of the titration, c) at the equivalence point. Answers are below, I need to know how to get to them step by step/conceptually Confirm: a) 8.49 , b) 9.21 c) 10.96

A 10 m3 mixture solution contains molecules A, B and C in water at ambient temperature....

A 10 m3 mixture solution contains molecules A, B and C in water at ambient temperature. Calculate: (a) Each species mass concentration. (b) Their molar concentration. (c) Total mass concentration and molar concentration. (d) Mass and mole fraction. given: Vmix= 10 m3 mA= 0.3 kg , mB= 0.5 kg , mC= 0.4 kg , mH2O= 9.8 kg MWA= 50 , MWB= 250 , MWC= 10 , MWH2O= 18

Give a more general expression for the magnitude of the torque T. Rewrite the answer found in Part A in terms of the magnitude of the magnetic dipole moment of the current loop m.

Give a more general expression for the magnitude of the torque T. Rewrite the answer found in Part A in terms of the magnitude of the magnetic dipole moment of the current loop m. Define the angle between the vector perpendicular to the plane of the coil and the magnetic field to be ф, noting that this angle is the complement of angle θ in Part A. Give your answer in terms of the magnetic moment m, magnetic field B, and...

10) Aluminum doesn’t readily corrode because: a) its standard reduction potential is more positive than oxygen...

10) Aluminum doesn’t readily corrode because: a) its standard reduction potential is more positive than oxygen b) its standard reduction potential is more negative than oxygen c) it forms a dense coating of alumina, Al2O3, that slows oxidation. d) It is already oxidized. . 11) What best describes a glass? a) A transparent, crystalline insulator. b) A viscous liquid c) An amorphous solid d) The thermodynamic (lowest energy) state for a material . For the following reaction answer questions 13...

You will be using the equations… q(metal) = -q(water) and q = m x C x...

You will be using the equations… q(metal) = -q(water) and q = m x C x delta T Two trials have been done. The following procedure is used: The initial temperature of ~200 ml of room temperature water is measured, and ~15 grams of an unknown metal is heated in boiling water from which the initial temperature of the metal is measured. At this point the heated metal is carefully added to the room temperature water, and the temperature of...

A 2.500×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.500×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL. Calculate the molality of the salt solution. XNaCl = 2.506 X 10-2 m Calculate the mole fraction of salt in this solution. XNaCl...

A 2.800×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.800×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.2 mL . The density of water at 20.0∘C is 0.9982 g/mL. 1a) Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. 1b) Calculate the...

A 2.250×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.250×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.4 mL . The density of water at 20.0∘C is 0.9982 g/mL. A. Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. B. Calculate the...

A 2.700×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.700×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL. Part A Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. Part B...

The Henry's law constant for helium gas in water at 30°C is 3.70 x10-4 M/atm.

The Henry's law constant for helium gas in water at 30°C is 3.70 x10-4 M/atm. If the partial pressure of helium above a sample of water is o.650 atm, what is the concentration of helium in water? A. 5.69 x10-4 M B. 1.76 103 M C. 1.30 M vD. 2.41 x10-4 M

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