The answer is the most probable location of electrons in an atom
- 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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Calculate the mass of the solute <span>in the solution :
Molar mass KCl = </span><span>74.55 g/mol
m = Molarity * molar mass * volume
m = 0.9 * 74.55 * 3.5
m = 234.8325 g
</span><span>To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm</span>³<span> and complete with water up to the mark, then cover the balloon and finally shake the solution to mix
hope this helps!</span>
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