Answer:
16.56 g
Explanation:
Mass is the production of Volume and Density.
m = V. d = 6 × 2.76 = 16.56 g
Answer:
turgor pressure can be done in a lab or a self test.
turgor pressure is key to the plant’s vital processes. It makes the plant cell stiff and rigid. Without it, the plant cell becomes flaccid. Prolonged flaccidity could lead to the wilting of plants.
Turgor pressure is also important in stomate formation. The turgid guard cells create an opening for gas exchange. Carbon dioxide could enter and be used for photosynthesis. Other functions are apical growth, nastic movement, and seed dispersal.
Explanation:
- salt is bad for turgor pressure.
- Turgidity helps the plant to stay upright. If the cell loses turgor pressure, the cell becomes flaccid resulting in the wilting of the plant.
- The wilted plant on the left has lost its turgor as opposed to the plant on the right that has turgid cells.
Answer:
1. 505g is the mass of the aluminium.
2. The answer is in the explanation
Explanation:
1. To solve this question we need to find the volume of the rectangle. With the volume and density we can find the mass of the solid:
Volume = 7.45cm*4.78cm*5.25cm
Volume = 187cm³
Mass:
187cm³ * (2.702g/cm³) = 505g is the mass of the aluminium
2. When the temperature of a liquid increases, the volume increases doing the density decreases because density is inversely proportional to volume. And works in the same way for gases because the temperature produce more collisions and the increasing in volume.
Answer:
c =0.2 J/g.°C
Explanation:
Given data:
Specific heat of material = ?
Mass of sample = 12 g
Heat absorbed = 48 J
Initial temperature = 20°C
Final temperature = 40°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 40°C -20°C
ΔT = 20°C
48 J = 12 g×c×20°C
48 J =240 g.°C×c
c = 48 J/240 g.°C
c =0.2 J/g.°C
