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3 years ago

All of the following are major pollutants of ground water except

Chemistry

1 answer:

Bogdan [553]3 years ago

6 0

All the following are major pollutants of ground water except CHLORINE FROM DRINKING WATER.

Ground water refers to the water source underneath the earth surface. This water source is very important to humans and plants and it is used for various purposes. The ground water source can be contaminated by pollutants such as fertilizers, pesticides, chemicals, radioactive wastes, etc.

Chlorine from drinking water can not contaminate ground water source, this is because, the amount of chlorine added to drinking water is very small and quite safe for human consumption. Apart from this, the chlorine usually dissipate from water few hours after its addition. Thus chlorine from drinking water does not pollute ground water sources.

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Which of these processes take place at the same temperature?

Elodia [21]

Answer:

the person above me is very right

Explanation:

7 0

2 years ago

What is the maximum mass of pure gold that could be extracted from 3.0kg of calaverite, a gold ore with the chemical formula AuT

Marina86 [1]

Answer:

1.3 × 10³ g

Explanation:

Step 1: Convert the mass of calaverite to grams

We will use the relationship 1 kg = 1,000 g.

$3.0kg \times \frac{1,000g}{1kg} = 3.0 \times 10^{3} g$

Step 2: Establish the appropriate mass ratio

The mass ratio of AuTe₂ to Au is 452.17:196.97.

Step 3: Calculate the maximum mass of pure gold that can be obtained from 3.0 × 10³ g of calaverite

$3.0 \times 10^{3} g Calaverite \times \frac{196.97gAu}{452.17gCalaverite} = 1.3 \times 10^{3} gAu$

3 0

3 years ago

At 800 K, the equilibrium constant, Kp, for the following reaction is 3.2 × 10–7. 2 H2S(g) ⇌ 2 H2(g) + S2(g) A reaction vessel a

djverab [1.8K]

Answer:

0.008945 atm

Explanation:

In the reaction:

2H2S(g) ⇌ 2 H2(g) + S2(g)

Kp is defined as:

$Kp = P_{H_{2}}^2P_{S_{2}} / P_{H_{2}S}^2$

<em>Where P is the pressure of each compound in equilibrium.</em>

If initial pressure of H2S is 3.00atm, concentrations in equilibrium are:

H2S = 3.00 atm - 2X

H2 = 2X

S2: = X

Replacing:

$3.2x10^{-7} = (2X)^2X / (3-2X)^2$

$3.2x10^{-7} = 4X^3 / 9- 6X+4X^2$

0 = 4X³ - 1.28x10⁻⁶X² + 1.92x10⁻⁶X - 2.88x10⁻⁶

Solving for X:

X = 0.008945 atm

As in equilibrium, pressure of S2 is X, <em>pressure is 0.008945 atm</em>

7 0

3 years ago

How many moles are in the number of molecules below? I only need to know the 5th question.

Arada [10]

Answer:

1. 6.02×10 23

this is the answer Hope it helps you

4 0

2 years ago

Calculate the Δ H°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. Δ H°f [CaCO3(s)] = –1206.9

maksim [4K]

Answer:

$\boxed{\text{178.3 kJ/mol}}$

Explanation

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

CaCO₃(s) ⟶ CaO(s) + CO₂(g)

ΔH°f/kJ·mol⁻¹: -1206.9 -635.1 -393.5

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ}& = & [(-635.1 + (-393.5)] - (-1206.9)\\& = & -1028.6 +1206.9\\& = & \mathbf{178.3}\\\end{array}\\\text{The enthalpy of decomposition is } \boxed{\textbf{178.3 kJ/mol}}$ l}}

7 0

2 years ago