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stira [4]
4 years ago
15

All of the following are major pollutants of ground water except

Chemistry
1 answer:
Bogdan [553]4 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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1. Using the Hoffman apparatus for electrolysis, a chemist decomposes 50.5 g of water into its gaseous elements. How many grams
Westkost [7]

Answer:

is this an experiment?

Explanation:

6 0
3 years ago
At about what age does basal metabolic rate begin to gradually decrease?
natulia [17]
This would decrease by 25 hope this helps !!
6 0
3 years ago
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In a titration experiment, how many moles of naoh will be required to completely neutralize 1 mole of nitric acid?
katrin [286]

Answer:

One mole

Explanation:

The balanced chemical equation is

<u>1</u>NaOH + <u>1</u>HNO₃ ⟶ NaNO₃ + H₂O

<u>1</u> mol       <u>1</u> mol

The coefficients in front of the formulas tell you the amount of something that reacts with an equivalent amount of something else.

In this reaction, 1 mol NaOH reacts with 1 mol HNO₃.

4 0
3 years ago
Read 2 more answers
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field
maksim [4K]

Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

7 0
3 years ago
What volume will 0.405 g of krypton gas occupy at STP?
Rufina [12.5K]

Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

                  83.8 g of Kr -------------------- 1 mol

                     0.405 g     -------------------  x

                     x = (0.405 x 1) / 83.8

                     x = 0.0048 moles

2.- Use the Ideal gas law to solve this problem

                   PV = nRT

- Solve for V

                      V = nRT / P

- Substitution

                      V = (0.0048)(0.082)(273) / 1

- Simplification

                       V = 0.108 / 1

- Result

                       V = 0.108 L

8 0
3 years ago
Read 2 more answers
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