3 years ago

What is the time required for contamination source

Chemistry

1 answer:

Afina-wow [57]3 years ago

8 0

Hi there!

It can take years to remove all of the harmful substances from the water.

I hope my answer helped :)

You might be interested in

Energy from a reaction was determined to be 3870 J using a water calorimeter, when burning 9.00 grams of H2. What is the total

Paraphin [41]

Answer:

860 J / mol

Explanation:

Enthalpy = Energy / no. of moles

no. of moles = mass / molar mass

Take the atomic mass of H = 1.0,

molar mass of H2 = 1.0 x 2

= 2.0

no. of moles of H2 = 9.00/2

= 4.5 mol

Hence,

Enthalpy = 3870 / 4.5

=860 J / mol

7 0

4 years ago

What is the boiling point for water in degrees celsius?

soldier1979 [14.2K]

Answer:

<h2>100°C is the boiling point of water in degrees Celsius</h2>

4 0

3 years ago

PLEASE HELP MY TEACHER IS GOING TO SEE ME AFTER CLASS

Crazy boy [7]

Answer:

The Nucelous helps prevent bad cells in

Explanation:

6 0

3 years ago

A sample of Cd(OH)2 is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of Cd +2 = 1.7 x 10 -5M

Archy [21]

Answer:

$Ksp=2.0x10^{-14}$

Explanation:

Hello there!

In this case, given the solubilization of cadmium (II) hydroxide:

$Cd(OH)_2(s)\rightleftharpoons Cd^{2+}(aq)+2OH^-(aq)$

The solubility product can be set up as follows:

$Ksp=[Cd^{2+}][OH^-]^2$

Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:

$Ksp=(1.7x10^{-5})(3.4x10^{-5})^2\\\\Ksp=2.0x10^{-14}$

Regards!

3 0

3 years ago

If you have a voltage of 120 volts entering your house and a light bulb with 0.625 amperes flowing through it, how many watts is

Luba_88 [7]

Answer:

75 watts

Explanation:

4 0

3 years ago

Read 2 more answers