- 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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I think it’s Carbon dioxide
I need help on this one too I think it might be (b)
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
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Answer:
1) 0.18106 M is the molarity of the resulting solution.
2) 0.823 Molar is the molarity of the solution.
Explanation:
1) Volume of stock solution = 
Concentration of stock solution = 
Volume of stock solution after dilution = 
Concentration of stock solution after dilution = 
( dilution )
0.18106 M is the molarity of the resulting solution.
2)
Molarity of the solution is the moles of compound in 1 Liter solutions.
Mass of potassium permanganate = 13.0 g
Molar mass of potassium permangante = 158 g/mol
Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)
0.823 Molar is the molarity of the solution.
