What is the importance of a special habitat?

How many moles of electrons must be transferred to plate out 110 g of manganese (MW ~ 55g/mol) from a solution of permanganate (

ololo11 [35]

Answer:

14 mol e⁻

Explanation:

Step 1: Write the balanced half-reaction for the reduction of permanganate to manganese

8 H⁺(aq) + 7 e⁻ + MnO₄⁻(aq) ⇒ Mn(s) + 4 H₂O(l)

Step 2: Calculate the moles corresponding to 110 g of manganese

The molar mass of Mn is 55 g/mol.

110 g × 1 mol/55 g = 2 mol

Step 3: Calculate the number of moles of electrons needed to produce 2 moles of Mn

According to the half-reaction, 7 moles of electrons are required to produce 1 mole of Mn.

2 mol Mn × 7 mol e⁻/1 mol Mn = 14 mol e⁻

8 0

2 years ago

Can an ion be either positive or negative? True or False

vekshin1

Answer:

True

Explanation:

3 0

2 years ago

Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for

Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

Hope this helps

4 0

2 years ago

Below are shown, for five metals, reduction reactions and standard electrode potential values. which of these metals is the leas

just olya [345]

Above question is incomplete. Complete question is attached below
........................................................................................................................
Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.

In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.

On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.

7 0

3 years ago

The law of universal gravitation describes the gravitational force between two objects.

Sliva [168]

Answer:

mass and distance

Explanation:

The strength of this force depends on the <u>mass</u>

of each object and

the <u>distance</u>

between them.

6 0

1 year ago