Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
A strong base turns phenolphthalein pink, so the answer is (3) NaOH
Answer:
0.190 M
Explanation:
Let's consider the neutralization reaction between HCl and NaOH.
HCl + NaOH = NaCl + H2O
11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:
0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol
The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.
1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:
M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M
Answer:
The pressure 10,000 m below the surface of the sea is 137.14 MPa.
The density 10,000 m below the surface of the sea is 2039 kg/m3
Explanation:
P0 and ρ0 are the pressure and density at the sea level (atmosferic condition). As the depth of the sea increases, both the pressure and the density increase.
We can relate presure and density as:
Rearranging
With this equation, we can calculate P at 10,000 m below the surface:
The density at 10,000 m below the surface of the sea is
Answer:
Incomplete question
Complete question:
8. Compartments A and B are separated by a membrane that is permeable to K+ but not to Na+ or Cl-. At time zero, a solution of KCl is poured into compartment A and an equally concentrated solution of NaCl is poured into compartment B. Which would be true once equilibrium is reached?
A. The concentration of Na+ in A will be higher than it was at time zero.
B. Diffusion of K+ from A to B will be greater than the diffusion of K+ from B to A.
C. There will be a potential difference across the membrane, with side B negative relative to side A.
D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.
E. The concentration of Cl- will be higher in B than it was at time zero.
Answer: D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.
Explanation:
Diffusion is the movement of molecules from region of higher concentration to lower concentration through a semipermeable membrane.
Since the k+ is the permeable membrane, the k+ ion in the KCl would move in equal magnitude and direction in the solution.
