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2 years ago

If a fish tank holds 6912 cubic inches of water, how many cubic feet of water are there in a fish tank.

Chemistry

1 answer:

stiv31 [10]2 years ago

6 0

Answer:

The answer is 576

Explanation:

There is 12 inches in a foot, divide the amount of inches by 12 to get your answer.

6912/12= 576

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Answer:

contact force

Explanation:

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2 years ago

54578 meters per second into miles per hour

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Answer:

54578 mps - 122087.51 mph

Explanation:

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2 years ago

Consider the two facts below:

OLEGan [10]

Answer:

A. There is more dissolved oxygen in colder waters than in warm water.

D. If ocean temperature rise, then the risk to the fish population increases.

Explanation:

Conclusion that can be drawn from the two facts stated above:

*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.

*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.

*Fishes, therefore, would thrive best in colder waters than warmer waters.

The following are scenarios that can be explained by the facts given and conclusions arrived:

A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)

D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).

4 0

2 years ago

Which physical property does the inventor need to consider when choosing a

adelina 88 [10]

Answer:

............

Explanation:

Ductility

6 0

3 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago