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3 years ago

How does temperature, pressure, and density impact the core of the Earth?

Chemistry

1 answer:

Vanyuwa [196]3 years ago

4 0

Answer:

temperatures within the outer core range from 7,200 to 9,000 F. Pressure also increases in the outer core due in part to the weight of the crust and mantle above.The Earth's magnetic field is generated by the outer core.

Explanation:

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Which is the most concentratic acid

denis23 [38]

Answer:

Tetraoxosulphate vi acid (H2SO4)

Explanation:

Becauses it ionizes completely

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3 years ago

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3 years ago

What is the difference between concentrated acid and a weak one?

makkiz [27]

Answer:

Difference between concentrated acid and weak acid :---

According to Arrhenius's theory the substances which easily get dissociated into H+ ions when dissolved in water are acids.

And the substance which readily gives H+ ions on dissociation (when dissolved in water) are Strong Acid. Examples are HCl , H2SO4 etc.

While,

Concentration of acid is just the value of pH. pH is the measurement of concentration of acid or base. The lower the pH, higher the concentration.

So strong acid is strong because it gives H+ ions readily in water and Concentration is the value of pH.

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3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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3 years ago

How moles are there in 15 grams of SiO2

BigorU [14]

The answer you’re looking for is 0.250

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3 years ago