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4 years ago

What interpretation can be made from an igneous rock body (such as granite) that has baked or metamorphosed the adjacent rocks?

Chemistry

1 answer:

Svetach [21]4 years ago

3 0

Answer:

Igneous rocks are constructed due to the cooling and solidification of molten magma. For example, Granite, Basalt, and Diorite. This solidification or crystallization process can take place below the earth's surface or at the earth's surface. These are often known as the intrusive (when magma crystallizes at a certain depth) and the extrusive (when crystallizes at the surface) igneous rocks.

Before the crystallization of magma and turning into granitic rocks, the molten magma present below the earth's surface, heats up the surrounding country rock, resulting in the alteration of its constituent minerals and rocks. When this radiated heat from the magma is extremely high, then it slowly leads to the changing of a rock type into metamorphic rocks.

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The normal freezing point of a certain liquid

slavikrds [6]

Answer : The molal freezing point depression constant of liquid X is, $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X = $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

$0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}$

$K_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of liquid X is, $4.12^oC/m$

3 0

3 years ago

If I have 340 mL of a 1.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it?

ipn [44]

Answer:

0.5667 M ≅ 0.57 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 1.5 M, V before dilution = 340 mL.

M after dilution = ??? M, V after dilution = 340 mL + 560 mL = 900 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (1.5 M)(340 mL)/(900 mL) = 0.5667 M ≅ 0.57 M.

5 0

3 years ago

By using the following reactions, calculate the heat of combustion of pentane:

Sergio [31]

To answer this problem, we use Hess' Law to calculate the overall enthalpy of the reactions. The goal is to add all the reactions such that the final reaction is C<span>5H12 (g) + 8O2 (g) → 5CO2 (g) + 6H2O (l) through cancellation adn multiplication. The first equation is multiplied by 5, the second one is multiplied by 6 and the third one is reversed. The final answer is -3538 J or -3.54 x10^3 kJ.</span>

6 0

4 years ago

A helium-filled weather balloon has a volume of 512 L at 18.9°C and 756 mmHg. It is released and rises to an altitude of 2.14 km

vazorg [7]

633.97 L

Explanation:

Well use the combined gas law;

P₁V₁T₁ = P₂V₂T₂

We need to change the temperatures into Kelvin;

18.9°C= 292.05 K

5.9°C = 279.05 K

756 * 512 * 292.05 = 639 * V₂ * 279.05

113,044,377.6 = 178,312.95 V₂

V₂ = 113,044,377.6 / 178,312.95

V₂ = 633.97 L

3 0

3 years ago

Read 2 more answers

Which of the following is NOT a chemical change?

nataly862011 [7]

Answer:

Melting butter

Explanation:

You can reverse the change of butter back to its original state but you can never reverse the rest back to there original state

4 0

3 years ago