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3 years ago

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A 3 L pocket of air at sea level has a pressure of 100 atm. Suppose the air pockets rise in the atmosphere to a certain height a

nd expands to a volume of 10 L. What is the pressure of the air at the new volume?

1 answer:

liraira [26]3 years ago

5 0

Answer:

30 atm

Explanation:

Initial Volume, V1 = 3L

Initial Pressure, P1 = 100 atm

Final Volume V2 = 10 L

Final Pressure, P2 = ?

These quantities are related by the Boyle's law equation which is given as;

V1P1 = V2P2

P2 = V1 * P1 / V2

P2 = 3 * 100 / 10

P2 = 30 atm

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Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

GarryVolchara [31]

Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

$T_{final}=25^oC$

Thus, the final temperature of the metal block is, $25^oC$

6 0

3 years ago

A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

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3 years ago

Formula of a copper (II)sulfate hydrate lab

s344n2d4d5 [400]

Answer:

Weigh the empty crucible, and then weigh into it between 2 g and 3 g of hydrated copper(II) sulphate. Record all weighings accurate to the nearest 0.01 g.

Support the crucible securely in the pipe-clay triangle on the tripod over the Bunsen burner.

Heat the crucible and contents, gently at first, over a medium Bunsen flame, so that the water of crystallisation is driven off steadily. The blue colour of the hydrated compound should gradually fade to the greyish-white of anhydrous copper(II) sulfate. Avoid over-heating, which may cause further decomposition, and stop heating immediately if the colour starts to blacken. If over-heated, toxic or corrosive fumes may be evolved. A total heating time of about 10 minutes should be enough.

Allow the crucible and contents to cool. The tongs may be used to move the hot crucible from the hot pipe-clay triangle onto the heat resistant mat where it should cool more rapidly.

Re-weigh the crucible and contents once cold.

Calculation:

Calculate the molar masses of H2O and CuSO4 (Relative atomic masses: H=1, O=16, S=32, Cu=64)

Calculate the mass of water driven off, and the mass of anhydrous copper(II) sulfate formed in your experiment

Calculate the number of moles of anhydrous copper(II) sulfate formed

Calculate the number of moles of water driven off

Calculate how many moles of water would have been driven off if 1 mole of anhydrous copper(II) sulfate had been formed

Write down the formula for hydrated copper(II) sulfate.

#*#*SHOW FULLSCREEN*#*#

Explanation:

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3 years ago

In the figure below, two nonpolar molecules are interacting.Which interaction would most likely cause these molecules to repel e

Fed [463]

Two non-polar molecules are most likely to interact by
induced dipole-induced dipole interaction.

Non-polar substances do not have a permanently established charge distribution due to similar electron affinities of the atoms that are present. Moreover, due to the absence of a polar hydrogen, they cannot exhibit hydrogen bonding. They interact with one another by induced dipole-induced dipole interactions which arise from the molecules of the substances coming into close vicinity of one another.

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3 years ago

Read 2 more answers

A high pH value indicates _____.

Mumz [18]

Answer:

A high pH value indicates a high concentration of OH- ion

Explanation:

The higher the OH- ion concentration high will be the pH.In simple words if the concentration of OH- ions are increased then the pH of the solution will also increase which means the solution will turns towards basic with increasing its OH- ion concentration.

Let us assume that the OH- concentration of a solution is 10-9 so the pOH of that solution will be 9 and the pH will be 5.

Now the concentration of OH-ion of that solution is increased from 10-9 to 10-8 now the pOH of that solution is 8 and the pH is 6.

3 0

3 years ago