- The independent variable (IV) is the lemon juice mixture
- The dependent variable (DV) is the appearance of the green slime on the shower
- The control variable (CV) are time taken to spray, the amount of spray
- The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
- The control group (CG) is the side of the shower sprayed with water.
INDEPENDENT VARIABLE
- Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.
DEPENDENT VARIABLE:
- Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.
CONTROL VARIABLE:
- Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray
CONTROL GROUP
- Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.
EXPERIMENTAL GROUP:
- Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.
Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:
- The independent variable (IV) is the lemon juice mixture
- The dependent variable (DV) is the appearance of the green slime on the shower
- The control variable (CV) are time taken to spray, the amount of spray
- The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
- The control group (CG) is the side of the shower sprayed with water.
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Explanation:
1. Sedimentation and decantation cannot be used for all types of mixtures.
Decantation is a separation technique in which is used to separate immiscible liquids or mixtures containing liquid and solids within them.
In decantation, gravity is used to bring the denser materials to settle at the bottom.
For homogenous mixtures, it is not possible to use decantation. A solution of sugar and water will not decant.
2. Yes, mass of an object reduces the settling time of such object in a mixture.
The higher the mass, the faster the rate of settling. Also, as we know, mass is directly proportional to density. A body with a high density will settle faster in solution.
Answer:
The new temperature of the water bath 32.0°C.
Explanation:
Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)
Initial temperature of the water = 
Final temperature of the water = 
Specific heat capacity of water under these conditions = c = 4.18 J/gK
Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J
( 1kJ=1000 J)




The new temperature of the water bath 32.0°C.
Answer :
The number of bonding pairs of electrons around the hydrogen atom = 2
The number of lone pairs of electrons around the hydrogen atom = 0
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that carbon has '4' valence electrons, hydrogen has '1' valence electrons and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in
= 1 + 4 + 5 = 10
According to Lewis-dot structure we conclude that, there are 8 number of bonding electrons and 2 number of non-bonding electrons.
The number of bonding pairs of electrons around the hydrogen atom = 2
The number of lone pairs of electrons around the hydrogen atom = 0
B. White Dwarf.
<h3>Explanation</h3>
The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.
As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:
Chandrasekhar Limit:
.
A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its
, which is much smaller than the Chandrasekhar Limit.
As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.