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bonufazy [111]
3 years ago
14

What kinds of problems or projects would a civil engineer work on?

Engineering
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

simple projects bovonhztisgx

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Explanation:

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Select four items that an industrial engineer must obtain in order to practice in the field.
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Professional engineering license

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If you want to obtain a Class E license before you turn 18, you must______. A. be in compliance with school attendance requireme
Makovka662 [10]

Answer:

The correct option is;

A. be in compliance with school attendance requirements

Explanation:

The requirements to acquire a special restricted driver's license for driver's license for under under 18 that have had a beginner's for up to 180 days include a minimum of 40 hours driving practice 10 of which should be in the dark. The application for the conditional drivers license is to be signed by the parent or guardian and the application is to be accompanied with proof of acceptable school attendance

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7 0
3 years 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

= \frac{moles}{Volume}

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

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

4 0
3 years ago
If an older multimeter is set to display its reading in megohms, what must you do to determine the correct value? Please help.
Gekata [30.6K]

Answer:

idk

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