Answer:
Answer is 'B' I think
Explanation:
Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Through previous experiments of shooting alpha particles, Rutherford knew they had considerable mass and speed.
A mixture is said to be homogeneous if its composition is uniform throughout the mixture. They are referred to as solutions. On the other hand, heterogeneous mixture does not have uniform composition. The substances present in the mixture have visible difference or phases.
Sea water is composed of salt, sand and water. Here, salt and water form homogeneous mixture but due to the presence of sand, the mixture is heterogeneous. Salt dissolves in water and the solution has uniform composition but sand does not dissolve in water, even after vigorous mixing, after some time it settles at the bottom resulting formation of layers. Thus, sea water containing salt, water and sand is a heterogeneous mixture.
We write DE = q+w, where DE is the internal energy change and q and w are heat and work, respectively.
(b)Under what conditions will the quantities q and w be negative numbers?
q is negative when heat flows from the system to the surroundings, and w is negative when the system does work on the surroundings.
As an aside: In applying the first law, do we need to measure the internal energy of a system? Explain.
The absolute internal energy of a system cannot be measured, at least in any practical sense. The internal energy encompasses the kinetic energy of all moving particles in the system, including subatomic particles, as well as the electrostatic potential energies between all these particles. We can measure the change in internal energy (DE) as the result of a chemical or physical change, but we cannot determine the absolute internal energy of either the initial or the final state. The first law allows us to calculate the change in internal energy during a transformation by calculating the heat and work exchanged between the system and its surroundings.
<span> </span> <span>V = nRT/P =
(0.875)(0.082057)(273)/(1) = 19.6 L</span>
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
