Answer:
D.phototropism
Explanation:
Phototropism is a type of tropism in which a plant or plant part responds to light. According to this question, a student wanted to investigate the effect of light on the growth of cress seedlings. The student used three different pots for the experiment. 
Pot 1 was placed with light from above. Pot 2 was placed in a cupboard with no light. Pot 3 was placed in a window with light from one direction only. However, the image attached to this question shows that the plants in the different pots face different directions in response to light, which depicts phototropism
 
        
             
        
        
        
There would be 79 electrons present in each atom of gold. I hope this helps :)
        
             
        
        
        
<h3>
Answer:</h3>
 2.47 × 10^24 molecules
<h3>
Explanation:</h3>
One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23. 
That is, 1 mole of a compound =  6.022 × 10^23 molecules
Therefore,
1 mole of Na₂CO₃ = 6.022 × 10^23 molecules
Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃
we get,
  = 4.1 moles × 6.022 × 10^23 molecules
  = 2.47 × 10^24 molecules
Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules
 
        
        
        
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit. 
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:  . .
- The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:  . .
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
 .
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be  of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio
 of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio  :
:
 .
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
 .
.