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2 years ago

How much gold is harvested per ton of earth from the underground mine?

1 answer:

ozzi2 years ago

5 0

If you are referring to NOVA - Hunting the Elements documentary, then about one ounce per ton of earth from the underground mine. One haul equals to about 400 tons, which means there are about 25 pounds of gold per every truck, and one ounce costs $1,800.

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In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00 degrees C. After

igomit [66]

Answer:

+26.6kJ/mol

Explanation:

The enthalpy of dissolution of NH₄NO₃ is:

NH₄NO₃(aq) + ΔH → NH₄⁺ + NO₃⁻

Where ΔH is the heat of reaction that is absorbed per mole of NH₄NO₃,

The moles that reacts in 1.60g are (Molar mass NH₄NO₃:80g/mol):

1.60g * * (1mol / 80g) = 0.02 moles reacts

To find the heat released in the coffee cup calorimeter, we must use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

1.60g + 75g = 76.60g

Change in temperature is:

25.00°C - 23.34°C = 1.66°C

Replacing:

Q = m×ΔT×C

Q = 76.60g×1.66°C×4.18J/g°C

Q = 531.5J

This is the heat released per 0.02mol. The heat released per mole (Enthalpy change for the dissolution of NH₄NO₃) is:

531.5J / 0.02mol = 26576J/ mol =

+26.6kJ/mol

<em>+ because the heat is absorbed, the reaction is endothermic-</em>

7 0

2 years ago

How are a carrot, an amoeba, and a bacterium alike?

PilotLPTM [1.2K]

Answer:

They all are made up of cells

8 0

2 years ago

A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Ca

umka21 [38]

Answer:

$W = 6.65 \cdot 10^{6} lbf$

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

$W = m\cdot g$ (1)

Where m the mass and g is the gravity

Also, we have that the mass is:

$m = \rho*V$ (2)

Where ρ is the density and V the volume

We cand calculate the volume as follows:

$V = L*w*d$ (3)

Where L is the length, w is the wide and d is the depth

By entering equation (2) and (3) into (1) we have:

$W = \rho*L*w*d*g$

$W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}$

$W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf$