Ask question

Ask question

All categories

faltersainse [42]

3 years ago

12

a factory wants to build a big tank to store chemicals. The tank will be near a river. Which of the following is a possible resu

lt of building a chemical tank?

1 answer:

harkovskaia [24]3 years ago

3 0

It would be B) because if the tank was to leak it would be harmful

You might be interested in

The world speed record on water was set on October 8, 1978 by Ken Warby. If Ken drove his motorboat a distance of 1500 m in 8.10

maxonik [38]

Answer:

317.52 mi/hr

Explanation:

First convert Meters into miles as the answer is required in miles/ h

1000m = 0.62 mi

Now, convert second into hours

7.45s = 0.0001 hr

The speed of the boat would be

v = 0.62/0.0001

=317.52 mi/hr

3 0

3 years ago

In 40 words or fewer, summarize the connection between kinetic and thermal energy.<br> PLEASE REPLY

xz_007 [3.2K]

Answer:

Explanation:

kenetic is made for thermal things

8 0

3 years ago

Jonathan is studying solutions in his chemistry class. The teacher instructs the students to measure 2 tablespoons of salt into

bulgar [2K]

8 tablespoons of water.
2 x 4 = 8

I hope this helps :)

3 0

3 years ago

How is calculating amplitude

kykrilka [37]

It is same as calculating maths for math

3 0

3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago