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3 years ago

8

Los fenomenos fisicos son cambios permamentes en la materia?

1 answer:

alexgriva [62]3 years ago

5 0

Answer:

Explanation:

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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

Sometimes, during the DNA replication process, mistakes are made. These are called

wolverine [178]

Answer: IT'S MUTATION I TOOK THE TEST ITS RIGHT HOPE THIS HELPED :)

8 0

2 years ago

Read 2 more answers

A gas exerts a pressure of 0.62 atm. Convert this to kPa and mmHg. Be sure to show your work.

IRINA_888 [86]

Answer:

The answer to your question is 0.62 atm = 62.82 kPa = 471.2 mmHg

Explanation:

Data

P = 0.62 atm

P = ? kPa

P = ? mmHg

Process

1.- Look for the conversion factor of atm to kPa and mmHg

1 atm = 101.325 kPa

1 atm = 760 mmHg

2.- Do the conversions

1 atm ----------------- 101.325 kPa

0.62 atm ------------ x

x = (0,62 x 101.325) / 1

x = 62.82 kPa

1 atm ------------------ 760 mmHg

0.62 atm ------------ x

x = (0.62 x 760)/1

x = 471.2 mmHg

4 0

3 years ago

If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams

Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of $KCl_3$ = 2.0 moles

Molar mass of $O_2$ = 32 g/mole

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 2.0 moles of $KClO_3$ react to give $\frac{2.0}{2}\times 3=3.0$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g$

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

$PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}$

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = $214.0^oC=273+214.0=487K$

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

$\rho$ = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

$P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}$

$P=1.78atm$

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0

3 years ago

pleeeaaaseeeee answer asappppppppp!!!!!!!!!!! Explain the processes of evaporation and condensation using the image above

goldfiish [28.3K]

In the lake that the rivers lead, water molecules evaporate into the sky and form clouds. In the sky, these water droplets condense and form clouds that will eventually rain.

5 0

3 years ago

Read 2 more answers