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Rasek [7]
2 years ago
8

What are alluvial diamond fields?

Chemistry
1 answer:
hammer [34]2 years ago
3 0

Answer:

Alluvial diamond mining occurs in riverbeds and beaches, where thousands of years of erosion and natural forces such as wind, rain, and water currents wash diamonds from their primary deposits in kimberlite pipes to beaches and riverbeds. Some alluvial deposits are from long-ago rivers.

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Calculate the amount of energy , in Joules, required to raise the temperature of 15.5 g of liquid water from 0.00o C to 75.0 oC.
deff fn [24]

Answer:

10043.225 J

Explanation:

We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:

Mass (m) = 15.5g

Latent heat of fussion of water (L) = 334J/g

Heat (Q1) =..?

Q1 = mL

Q1 = 15.5 x 334

Q1 = 5177 J

Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.

This is illustrated below:

Mass = 15.5g

Initial temperature (T1) = 0°C

Final temperature (T2) = 75°C

Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C

Specific heat capacity (C) of water = 4.186J/g°C

Heat (Q2) =?

Q2 = MCΔT

Q2 = 15.5 x 4.186 x 75

Q2 = 4866.225 J

The overall heat energy needed is given by:

QT = Q1 + Q2

QT = 5177 + 4866.225

QT = 10043.225 J

Therefore, the amount of energy required is 10043.225 J

8 0
3 years ago
Why do some objects float why others sink
Leona [35]
Because so objects are denser than water and some are less dense than water
4 0
3 years ago
Read 2 more answers
The illustration to the left represents a mixture of hydrogen ( light blue ) and iodine ( purple ) reacting to form a product. I
Ksju [112]

Answer:

1. Hydrogen Iodide

2. 6 molecules of Hydrogen Iodide

3. Iodine is the limiting reagent

Explanation:

The image of the illustration in the question has been attached:

1. The illustration represents a mixture of hydrogen ( light blue ) and iodine ( purple )

H₂ + I₂ ---> 2HI

This forms hydrogen iodide.

2. In the given illustration, 6 product molecules of Hydrogen Iodide. This is indicated in the box on the right side of the illustration.

3.  The limiting reagent is the reactant that determines how much of the products are made. It is the substance that is totally consumed when the chemical reaction is completed. In the box on the right side of the illustration, you will see that hydrogen which is indicated by blue is in excess. The limiting reagent is the one that is completely consumed which is the iodine.

8 0
3 years ago
A fat is composed of long chains of carbon and hydrogen atoms. In a reaction with a strong base, a fat forms a soap and glycerol
Nonamiya [84]

Answer:

Empirical formula is C₉H₁₅O

Molecular formula = C₈₁H₁₃₅O ₉

Explanation:

Percentage of carbon = 77.60%

Percentage of oxygen = 11.45%

Percentage of hydrogen = 10.95%

Molecular weight = 1253 g/mol

Molecular formula = ?

Empirical formula = ?

Solution:

Number of gram atoms of C = 77.60 g /12g/mol =6.5

Number of gram atoms of O = 11.45 g / 16 g/mol = 0.72

Number of gram atoms of H = 10.95 g / 1.008 g/mol= 10.9

Atomic ratio:

C               :            H                 :    O

6.5/0.72   :       10.9/0.72         :   0.72/0.72

     9          :            15              :         1

C : H : O = 9 : 15 : 1

Empirical formula is C₉H₁₅O

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 1253 / 139

n = 9

Molecular formula = n (empirical formula)

Molecular formula = 9 (C₉H₁₅O )

Molecular formula = C₈₁H₁₃₅O ₉

4 0
3 years ago
The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates:
alexira [117]

Answer:

Kc = 6x10⁻⁶

Explanation:

For the reaction:

4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g)

Kc is defined as:

Kc =[N₂]² [H₂O]⁶ / [NH₃]⁴ [O₂]³

The equilibrium concentrations of the gases is -Because volume of the container is 1.00L-:

[N₂] = 2X = 1.96x10⁻³; <em>X = 9.8x10⁻⁴</em>

[H₂O] = 6X; 6ₓ9.8x10⁻⁴ = 5.88x10⁻³

[NH₃] = 0.0150M - 4X = 0.01108M

[O₂] = 0.0150M - 3X = 0.01206M

Replacing in Kc expression:

Kc =[1.96x10⁻³]² [5.88x10⁻³]⁶ / [0.01108M]⁴ [0.01206M]³

<h3>Kc = 6x10⁻⁶</h3>

8 0
3 years ago
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