In this experiment you will determine the density of an unknown liquid. You will measure the mass of a known volume delivered from a pipet and use the expression:
<span><span><span>M
V</span></span><span>D=</span><span /></span>
where D = density, M = mass and V = volume to calculate the density of the unknown liquid.
Answer:
the answer is 2.75 but since that isn't an option I'd go with 1 because I just took a test with this question and it was the right answer.. hopefully that helps
Answer:
Both b and d can be correct
Explanation:
Generally, diffusion does not require energy (<em>making option a wrong</em>) because it is the movement of particles from a region of high concentration to a region of low concentration hence diffusion moves particles in the direction of a concentration gradient. An example of this is the passive transport (for instance, uptake of glucose by a liver cell).
However, in some cases, when diffusion is against the concentration gradient (i.e when particles move from a region of low concentration to a region of high concentration), diffusion will require energy in a case like this (<em>making option c wrong</em>). An example of this is active transport (transport of protein called sodium-potassium pump which involves pumping of potassium into the cell and sodium out of the cell).
The explanation above shows that diffusion can require energy to move particles (in or out) of the cell through the cell membrane.
Answer:
A
Explanation:
I would say a because we all know that kinetic energy is energy in motion and fuel is an example of chemical energy. So it must be from chemical to kinetic energy.
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol