Answer:
C. Explain ONE way in which the USA & USSR competed against each other during the Cold
War in the Post-World War Two period (c. 1945-present)
HURRY
Dr. Khan works for the marketing department of a company that manufactures mechanical toy dogs. Dr. Khan has been asked to assess the effectiveness of a new advertising campaign that is designed to be most persuasive to people with a certain personality profile. She brought four groups of participants to the lab to watch the video advertisements and to measure the likelihood that they would purchase the toy, both before and after watching the ad. The results of Dr. Khan’s study are presented below.
Part A
Explain how each of the following concepts applies to Dr. Khan’s research.
Survey
Dependent variable
Big Five theory of personality
Part B
Explain the limitations of Dr. Khan’s study based on the research method used.
Explain what Dr. Khan’s research hypothesis most likely was.
Part C
Use the graph to answer the following questions.
How did the trait of agreeableness affect how people responded to the new ad campaign?
How did the trait of conscientiousness affect how people responded to the new ad campaign?
Answer:
35.9 ml
Explanation:
Start with the balanced equation:
3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)
This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-
∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4
We know that concentration = moles/volume i.e:
c= n/v
∴n=c×v
∴nCuCl2=0.107×91.01000=9.737×10−3
I divided by 1000 to convert ml to L
∴nNa3PO4=9.737×10−3×23=6.491×10−3
v=nc=6.491×10−30.181=35.86×10−3L
∴v=35.86ml
Answer:
Mass of carbon dioxide produced = 52.8 g
Explanation:
Given data:
Mass of carbon react = 14.4 g
Mass of oxygen = 56.5 g
Mass of oxygen left = 18.1 g
Mass of carbon dioxide produced = ?
Solution:
C + O₂ → CO₂
Number of moles of C:
Number of moles = mass/ molar mass
Number of moles = 14.4 g/ 12 g/mol
Number of moles = 1.2 mol
18.1 g of oxygen left it means carbon is limiting reactant.
Now we will compare the moles of C with CO₂.
C : CO₂
1 : 1
1.2 : 1.2
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.2 mol × 44 g/mol
Mass = 52.8 g
When dealing with making diluted solutions from concentrated solutions, we can use the following formula
c1v1 = c2v2
where c1 and v1 are the concentration and volume of the concentrated solution respectively.
c2 and v2 are the concentration and volume of the diluted solution respectively
substituting these values in the above formula,
20 mL x 0.200 M = C x 250.0 mL
C = 0.0160 M
