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2 years ago

Whats the density of mass?

Chemistry

1 answer:

Fudgin [204]2 years ago

7 0

Answer:

it depends on the mass of the object but sometimes smaller objects have more density the higher the mass the the lower the density

You might be interested in

What happens to an earthquake’s p-waves when they strike the liquid core?

padilas [110]

Answer:

They bounce off.

Explanation:

They change into S-waves. They change into R-waves.

6 0

3 years ago

A student places 1.38g of unknown metal at 99.6C into 60.50g of water at 22.1C. The entire system reaches a uniform temperature

stiv31 [10]

Explanation:

Heat energy lost by metal = Heat energy gained by water.

(0.00138kg) * c * (99.6-31.6) = (0.006050kg) * 4.148 * (31.6-22.1)

c = 2.541J/kg°C

8 0

3 years ago

A natural water with a flow of 3800 m3/d is to be treated with an alum dose of 60 mg/L. Determine the chemical feed rate for the

svet-max [94.6K]

Explanation:

First, we will calculate the feed rate of alum as follows.

$\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}$

= 228000 g/day

Converting this amount into g/min as follows.

$\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}$

= 158 g/min

Now, the chemical equation will be as follows.

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}$

= 0.151 mmol $mmol SO^{2-}_{4}/L$

$\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}$

= 15.15 mg $CaCO_{3}$ /L

For precipitate:

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}$

= 7.88 $Al(OH)_{3}/L$

$\frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}$

= 29.9 $Al(OH)_{3}/day$

3 0

3 years ago

4. The number of waves that pass the poster per second is called the of the <br> waves.

Law Incorporation [45]

Answer:

Explanation:

6 0

2 years ago

How would you expect a sample of pentane to behave under similar experimental conditions?

ExtremeBDS [4]

I would expect a sample of pentane to act the same under similar experimental conditions. It should follow the gas law conditions set. Not only that, it is also in the same family as the others so it is likely closer than other gases in numbers.

3 0

3 years ago