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4 years ago

Which is the most accurate description of ionic and covalent bonding?

1 answer:

6 0

Ionic Bonding:
This type of bonding occurs when atoms want to fulfil their valence shells by taking/giving electrons to other atoms. This, leads to completed valence shells in most cases and ionisation of both elements. The opposite charges cause the elements to stick together because opposites attract.

Covalent Bonding:
This type of bonding occurs when electrons are shared between atoms to each fill up their own valence shells by sharing. The balance between the attraction forces and repulsion forces between the shared electrons is called covalent bonding.

Hope I helped :)

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When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of

larisa86 [58]

The unbalanced chemical reaction is
Na (s) + H2O (g) yields NaOH (aq) + H2 (g)

By inspection,
There is 1 mole of Na in both sides
There is 2 moles of H in the left and 3 moles in the right side.
There is 1 mole of O in both sides

We balance the reaction by:

2Na (s) + 2H2O (g) yields 2NaOH (aq) + H2 (g)

The answer is 2

4 0

3 years ago

snow_tiger [21]

There are 1,000m is 1k. So just move the decimal one position right. 127.56m

There are 10,000cm in 1k. Move the decimal two positions right. 1275.6cm

5 0

4 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

What does every force have?

stealth61 [152]

A force has both size and direction.

5 0

3 years ago

Give 3 examples of solutions, suspensions and colloids..

Vlad [161]

Some examples of solutions":" are salt water, rubbing alcohol, and sugar dissolved in water.

examples of suspensions are muddy water milk of magnesia Slaked lime for whitewashing.

examples of colloids are whipped cream, mayonnaise, milk,

5 0

3 years ago

Read 2 more answers