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:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
where,
- c: specific heat capacity
- m: mass
- ΔT: change in the temperature
The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.
Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.
Answer:
The volume is increased.
Explanation:
According to <em>Charles' Law</em>, " <em>at constant pressure the volume and temperature of the gas are directly proportional to each other</em>". Mathematically this law is presented as;
V₁ / T₁ = V₂ / T₂ -----(1)
In statement the data given is,
T₁ = 10 °C = 283.15 K ∴ K = 273.15 + °C
T₂ = 20 °C = 293.15 K
So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,
V₁ = 100 L
V₂ = Unknown
Now, we will arrange equation 1 for V₂ as,
V₂ = V₁ × T₂ / T₁
Putting values,
V₂ = 100 L × 293.15 K / 283.15 K
V₂ = 103.52 L
Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.
Answer:
400 miles because it is going 50 mph
