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Dominik [7]
1 year ago
13

which of the following is not a general education elective area? group of answer choices humanities/fine arts social/behavioral

sciences critical thinking physical and life sciences/mathematics
Engineering
1 answer:
galben [10]1 year ago
8 0

Option D is correct. The component of college life known as general education deals with the knowledge, abilities, attitudes, and values that define educated people.

It respects the linkages between different bodies of knowledge and is not constrained by disciplines. Graduates with a NOVA degree will have the ability to write effectively and engage in civic participation, critical thinking, professional preparation, numeric literacy, and scientific literacy. As a result, the College has chosen the following list of general education electives in accordance with the general education standards of the Virginia Community College System. In order to choose the best course for their curriculum and/or transferability to another college, it is strongly advised that students speak with their academic advisor or counselor.

Learn more about system here-

brainly.com/question/19548032

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Which explanation best summarizes what went wrong during Paul’s cost analysis?
Valentin [98]

Answer:

wut is it

Explanation:

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1 year ago
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

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= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

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We simply applied the above formulas

4 0
3 years ago
A DC generator turns at 2000 rpm and has an output of 200 V. The armature constant is 0.5 V-min/Wb, and the field constant of th
WITCHER [35]

Answer:

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200=20×I

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4 0
3 years ago
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Answer:

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