Answer:
answer is gold
Explanation:
bcz gold is having a property of malleable so it can be drawn into thin sheets
Answer:
This question is incomplete, the remaining part of the question is:
What is the control group, independent variable and dependent variable?
Control group: Plants placed in 80 degree rooms
Independent variable: Change in temperature
Dependent variable: Change in color of leaves
Explanation:
The independent variable in a scientific experiment is the variable that the experimenter controls or manipulates in order to bring about a change in the dependent variable. In this experiment, the variable manipulated by Justin B is the TEMPERATURE CHANGE.
On the other hand, a variable is said to be dependent if it is the variable that responds to a change made to the independent variable or rather it is the outcome. In this experiment, Justin B is trying to see the outcome on the color change in leaves when exposed to a low temperature, hence, COLOR CHANGE IN LEAVES is the dependent variable.
Control group of an experiment is the group that receives no experimental treatment. It is the group the experimenter considers normal and hence is comparing with his experimental group. In this experiment, Justin B believes the leaves change color in a low temperature, hence, he placed some plants in a lower temperature (60 degree) in order to compare them with when the plants are placed in a higher temperature (80 degree). As far as this experiment is concerned, the plants placed in 80 degrees temperature are believed by Justin B not to undergo color change, hence, they are the CONTROL GROUP while the group he placed in 60 degrees temperature are what he is interested in, making them the EXPERIMENTAL GROUP
An aqueous solution of potassium sulfate exhibits colligative properties. Colligative properties are properties that depends on the concentration of a substance in a solution. These properties are freezing point depression, vapor pressure lowering, osmotic pressure and boiling point elevation. For this problem we use the concept of freezing point depression since we are given the freezing point of the solution. Freezing point depression is as:
ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
<span>m = 0.4014
Thus, the molality of the solution is 0.4014.</span>
It would be 4.6cgL not sure tho because I didint do that good in this
