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arlik [135]
2 years ago
7

In Fig C, which pair of electrons is the lone pair?

Chemistry
1 answer:
arlik [135]2 years ago
5 0

Answer:

bottom right top left fig

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How many moles of ammonia (nh3) would be produced if 2.5 moles of nitrogen (n2) reacted with excess hydrogen (h2)?
cluponka [151]
Firstly, a balanced equation has to be written for the production of ammonia (NH₃) from hydrogen gas (H₂) and nitrogen gas (N₂):
                    N₂   +   3H₂   →   2NH₃

Now, the mole ratio of N₂ : NH₃ is 1 : 2 based on the coefficients of the balanced equation.

If the moles of N₂ = 2.5 moles

   then the moles of NH₃ produced = 2.5 mol × 2 
                                                          =  5 mol

Thus, the moles of ammonia produced when 2.5 mol of nitrogen gas is combined with excess hydrogen gas is 5 mol.
6 0
3 years ago
Balance the following chemical equation I would give anyone the brainiest for answering.
serg [7]

<u>Answer :</u>

Part 13:

The balanced chemical reaction will be:

Cu(NO_3)_2+Na_2SO_4\rightarrow CuSO_4+2NaNO_3

Part 14:

The balanced chemical reaction will be:

Cu(NO_3)_2+Na_2C_2O_4\rightarrow CuC_2O_4+2NaNO_3

Part 15:

The balanced chemical reaction will be:

Cu(NO_3)_2+2KOH\rightarrow Cu(OH)_2+2KNO_3

<u>Explanation :</u>

Balanced chemical reaction : It is defined as the reaction in which an individual element of an atom present on reactant side must be equal to product side.

Part 13:

The balanced chemical reaction will be:

Cu(NO_3)_2+Na_2SO_4\rightarrow CuSO_4+2NaNO_3

Part 14:

The balanced chemical reaction will be:

Cu(NO_3)_2+Na_2C_2O_4\rightarrow CuC_2O_4+2NaNO_3

Part 15:

The balanced chemical reaction will be:

Cu(NO_3)_2+2KOH\rightarrow Cu(OH)_2+2KNO_3

5 0
3 years ago
Which observation does not indicate that a chemical reaction has occurred?
castortr0y [4]

Answer:

change in the total mass of substances

6 0
3 years ago
What is the transfer of electrons in Al + Cl = AlCl3
otez555 [7]

Answer:

3 e⁻ transfer has occurred.

Explanation

This is a redox reaction.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
  • [Ne]= (1s²) (2s² 2p⁶)

A combination of both the reactions( Half-reactions) leads to a redox reaction.

Let us look at initial configurations of Al and Cl

[Al]= 1s² 2s² 2p⁶ 3s² 3p¹

[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵

Hence, Al can lose 3 electrons to achieve octet config.

and, Cl can gain 1e to achieve nearest noble gas config. [Ar]

This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.

Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃

Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)

Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.

3 0
3 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

<h2>[OH⁻] = 14.29 </h2>

Hope this helps

4 0
2 years ago
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