Percentage yield=(actual yield/theoretical yield) x 100%
= (45/50) x 100%
= 90%
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?
Light is one form of energy that travels in electromagnetic waves. This energy is both magnetic and electrical. There are many different types of electromagnetic (EM) waves.
Answer:
pH = 4.543
Explanation:
- CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
- pKa = - Log Ka
∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
∴ pKa = 4.87
⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]
added 300 mL 0f 0.02 M NaOH:
⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)
⇒ <em>C</em> CH3CH2COOH = 0.048 M
⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M
mass balance:
⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)
charge balance:
⇒ [H3O+] + [Na+] = [CH3CH2COO-]
∴ [Na+] = 0.02 M
⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)
(2) in (1):
⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]
replacing in Ka:
⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])
⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]
⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]
⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0
⇒ [H3O+ ] = 2.867 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = 4.543
Wind speed bc The faster the wind, the longer it blows, or the farther it can blow uninterrupted, the bigger the waves.