All categories

3 years ago

Ionic bonds result from elements_____.

1 answer:

4 0

Ionic bonding is a type of chemical bond in which valence electrons are lost from one atom and gained by another. This exchange results in a more stable, noble gas electronic configuration for both atoms involved. An ionic bond is based on attractive electrostatic forces between two ions of opposite charge

You might be interested in

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

3 years ago

Barium and lead both produce yellow precipitate with the chromate ion as part of their confirmation tests. if you had a sample t

UkoKoshka [18]

Barium is an element that is primarily used in fireworks because of its distinct green colour while on the other hand, lead is used in electrical industries due to its unique properties. The ions of these elements can simply be identified when its insoluble salts get precipitated by any process.

5 0

3 years ago

Read 2 more answers

Find the percent composition of C7H5N3O6

Oksana_A [137]

Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.

First, let's find the total mass by using the masses of the elements given on the periodic table.

7 x 12.011 (mass of Carbon) = 84.077

5 x 1.008 (mass of Hydrogen) = 5.04

3 x 14.007 (mass of Nitrogen) = 42.021

6 x 15.999 (mass of Oxygen) = 95.994

Add all of those pieces together.

84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.

Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %

Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %

Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %

Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %

You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.

5 0

2 years ago

Hwlp meeeeeeeeeeeeeeeeeeee

Vanyuwa [196]

The most appropriate answer is C !!

4 0

3 years ago

3. 200 ml of 0.2 M HCl is neutralized with 0.1 M

algol [13]

Answer:

a. 0.1 M

Explanation:

By definition, half neutralization is the point at which half of the acid has been neutralized.

The neutralization reaction that you are studying is the acid-base reaction:

HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)

Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:

Half molarity = M / 2 = 0.2 M / 2 = 0.1 M

So, the answer is the first choice: a. 0.1 M.

3 0

3 years ago