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1 year ago

Scientific process how can a casual question be answered 3 hypotheses water experiment

Chemistry

1 answer:

Vlad1618 [11]1 year ago

3 0

There are 3 experiments which helps to under water evaporation system.

Both the temperature of the air surrounding the water and the rate of evaporation increased.
In the second experiment, we observed an increase in air flow around the water, and this time, the rate of evaporation increased as well.
In the third and final experiment, we observed an increase in the amount of light in the room, and this time, the rate of evaporation increased as well.
Because the kinetic energy of a substance can be measured by its temperature, increasing in the temperature or kinetic energy of the air around water will also increase the kinetic energy of the water, which will increase the rate of evaporation.
It's crucial to have good air flow since it drives water droplets in the air away from you, reducing the amount of humidity in the region. Because there will then be space for the evaporated air or space for the evaporated water to flow into the air, it will then be simpler for the water to really evaporate.
Water in the presence of sunlight is given extra energy, which increases evaporation.

Learn more about Evaporation here:

brainly.com/question/24258

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Which energy resource causes the most safety concerns?.

Vlad [161]

Answer:

Based on safety and carbon emissions, fossil fuels are the dirtiest and most dangerous, while nuclear and modern renewable energy sources are vastly safer and cleaner.

4 0

2 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

Now put all the given values in the above formula, we get:

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0

3 years ago

What may likely to happen with the snakes If the rats decrease in number?

lions [1.4K]

If the number of rats decrease, snakes would look for another prey.

5 0

2 years ago

Read 2 more answers

A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add

erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can use the formula C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

47.2 mL * 150 mM = 24.0 mM * V₂

V₂ = 295 mL

And converting into L becomes:

295 mL * $\frac{1 L}{1000mL}$ = 0.295 L

6 0

3 years ago

You are working in your laboratory and decide to do some cleaning. You find a test tube with some brown substance congealed at t

Anna [14]

Answer:

The mystery substance is C. Bromine (Br)

Explanation:

Argon (Ar) is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.

Scandium (Sc) is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.

Both bromine and iodine are halogens (group 17 of the periodic table).

The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.

The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.

Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.

You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.

Therefore, the mystery substance is bromine (Br).

7 0

3 years ago