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inysia [295]
3 years ago
5

Which statement supports the claim that the continental crust was once joined together as one supercontinent, Pangaea?

Chemistry
1 answer:
Liula [17]3 years ago
5 0

Answer:

each continent shifts 2-5 inches per year so once it was joined then it is what it is now.In a million years again it will comes closer and wjo knows what will happen.moutains will be formed as the tectonic plates will collide.

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Determine the mole fraction of the solute if you have 3.4 grams KCl in 8.1 grams H2O
Yakvenalex [24]
.42; 3.4 divided by 8.1 is about .42
6 0
3 years ago
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How traits are passed from parents to offspring
pantera1 [17]
Its genes from your parents 

6 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

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

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera
Grace [21]

Answer: The initial temperature of the iron was 515^0C

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 iron = 360 g

m_2 = mass of water = 750 g

T_{final} = final temperature = 46.7^0C

T_1 = temperature of iron = ?

T_2 = temperature of water = 22.5^oC

c_1 = specific heat of iron = 0.450J/g^0C

c_2 = specific heat of water= 4.184J/g^0C

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

-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]

T_i=515^0C

Therefore, the initial temperature of the iron was 515^0C

4 0
3 years ago
a gas in a container with a fixed volume was originally at a pressure of 317kpa and a temperature of 154k. what is the new press
goldfiish [28.3K]
<h2>Hello!</h2>

The answer is: 4.77atm

<h2>Why?</h2>

Since there's a fixed volume, we can use the the Gay-Lussac's Law which stablish a relation between the pressure and the temperature:

\frac{P}{T}=k

<em>P</em> is the volume of the gas

<em>t</em> is the temperature of the gas

<em>k </em>is the proportionality constant

We also have the following equation:

\frac{P1}{T1}=\frac{P2}{T2}

Where:

P2=\frac{P1*T2}{T1}=\frac{317kPa*235K}{154K}=483.73kPa

We are asked to find the pressure in atm, so we must convert 483.73kPa to atm:

1kPa=0.009869atm

Then,

483.733kPa*\frac{0.009869atm}{1kPa}=4.77atm

Have a nice day!

8 0
3 years ago
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