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iVinArrow [24]
2 years ago
11

Two tons of water is pumped into a tank 15 ft. High in 5 minutes. Determine the power output of the pump. (Note: One ton is equa

l to 2,000 lb. And remember to convert the minutes to seconds.)
Chemistry
1 answer:
Semmy [17]2 years ago
5 0

Answer:

200 lb•ft/s

Explanation:

From the question given above, the following data were obtained:

Force (F) = 2 tons

Time (t) = 5 mins

Height (h) = 15 ft

Power (P) =?

Next, we shall convert 2 tons to pound. This can be obtained as follow:

1 ton = 2000 lb

Therefore,

2 tons = 2 × 2000

2 tons = 4000 lb

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

5 mins = 5 × 60

5 mins = 300 s

Finally, we shall determine the power of the pump. This can be obtained as follow:

Force (F) = 4000 lb

Time (t) = 300 s

Height (h) = 15 ft

Power (P) =?

P = F × h / t

P = 4000 × 15 / 300

P = 60000 / 300

P = 200 lb•ft/s

Thus, the power of the pump is 200 lb•ft/s

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mamaluj [8]

Answer:

  • The first picture attached is the diagram that accompanies the question.

  • The<u> second picture attached</u> is the diagram with the answer.

Explanation:

In the box on the left there are 8  Cl⁻ ions and 8 Na⁺ ions.

The dissociaton equation for NaCl(aq) is:

  • NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)

The dissociation equation for CaCl₂ (aq)  is:

  • CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)

A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.

Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half)  and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).

Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.

4 0
3 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

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Answer: check explanation

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Mass of the chain= 3 kg/m × 5 m.

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