Imagine you have been asked to create an experimental design to test the hypothesis that talking on a cell phone impairs driving skills. explain why a control condition would be important to include in testing this hypothesis. how should subjects be assigned to conditions? how can the researchers design the experiment so that the only difference between both conditions is the use of a cell phone?

Explanation:

To see if using a cell phone actually impairs driving skills, it is necessary to conduct an experiment to test this hypothesis. We can define an experiment as a controlled test to affirm a certain behavior (hypothesis). This is an important procedure to discover the causes and effects that will be important to complete an experiment, so when performing this test it is necessary for a researcher to manipulate a variable in order to find out what its effects are on another variable, so that conclusions about cause and effect.

Every experiment must have an independent variable and a dependent variable. The independent variable is configured as the experimental situation manipulated by the researcher, and the dependent variable is defined as the behavior that the researcher tries to predict. In this question, then, the independent variable is the use of the cell phone, and the dependent variable is the number of collisions.

Therefore, to be valid, an experiment must have an experimental condition and a control condition.

(75°f − 32) × 5/9 = 23.889°c

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂