The type of energy used is kinetic energy. Kinetic energy is the energy of motion.
Given the volume of HCl solution = 30.00 mL
Molarity of HCl solution = 0.1000 M
Molarity, moles and volume are related by the equation:
Molarity = 
Converting volume of HCl from mL to L:

Calculating moles of HCl from volume in L and molarity:

The final moles would be reported to 4 sig figs. So the correct answer will be 0.03000 mol HCl
Correct option: C. 0.03000mol
Answer:.633
Explanation:
I have know idea but it was right
Sodium potassium pump is an active pump which transfer sodium and potassium ions across the membrane with the expenditure of energy in the form of ATP.
This kind of pump is generally used in nerve cells.
The pump works against the concentration gradient as the pump moves three Na+ ions outside the cell and two K+ ions inside the cell, though there is a high concentration of Na+ outside the cell and a low concentration of K+ outside the cell.
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5