2. You water three sunflower plants with salt water. Each plant receives a different concentration of salt solutions. A fourth plant receives pure water. After a two week period, the height is
measured.
IV the amount of salt in the water
DV the height
CG plant that gets just water

This question is asking to identify the variables in the experiment

Independent variable: different salt concentration

Dependent variable: Height of plants

Control group: Plants that receive pure water

Experimental group: Plants that receive different concentration of the salt water solution

Explanation:

The independent variable is the variable that is manipulated or changed by the experimenter in order to bring about a measurable outcome. In this experiment, the independent variable is the DIFFERENT SALT CONCENTRATION the plants were exposed to.

The dependent variable is the variable that responds to the change or manipulation of the independent variable or the measured variable. In this experiment, the dependent variable is the HEIGHT OF THE PLANTS.

The experimental group refers to the group that receives the experimental treatment. The experimental group in this experiment are the PLANTS THAT RECEIVE DIFFERENT CONCENTRATION OF SALT SOLUTIONS.

The control group does not receive any experimental treatment as the independent variable is unchanged. The control group is the PLANT THAT RECEIVE PURE WATER i.e no salt solution

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol