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3 years ago

What is the difference between a molecular formula, structural formula and an electron dot formula? Give an example of each

Chemistry

1 answer:

jeka943 years ago

7 0

Answer:

The molecular formula tell us what elements the atoms are, and how many moles and atoms are attributed toward each element. For example, molecular formula of glucose is . That means one molecule of glucose has 6 molecules of C, 12 molecules of hydrogen and 6 molecules of oxygen.

The structural formula of a chemical compound is a graphic representation of the molecular structure.

Electron dot diagram or a Lewis diagram or a Lewis structure.

Explanation:

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Why does a reaction slow down with time?

nignag [31]

Answer:

D. The reactant concentrations decrease.

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3 years ago

3h2so4 2b(oh)3 - b2(so4)3 6h2o what is the mole ratio between water and sulfuric acid?

Nutka1998 [239]

Since there is an equal number of each element in the reactants and products of 3H2SO4 + 2B(OH)3 = B2(SO4)3 + 6H2O, the equation is balanced.

<h3>What is Sulfuric acid?</h3>

Sulfuric acid (H2S04) is a corrosive substance, that destroys the skin, eyes, teeth, and lungs. Severe vulnerability can result in death. Workers may be harmed from disclosure to sulfuric acid. The level of exposure depends on

The sulfuric acid bleaching can substitute the presently assumed oxygen and chlorine stages if the additives are allowed. This bleaching process was also practical for oxygen-bleached hardwood kraft pulp, but it was less compelling for softwood kraft pulp and oxygen-bleached softwood kraft pulp.

Disbanding sulfuric acid in water is an exothermic procedure. When sulfuric acid is mixed with ice this exothermic procedure causes the temperature to rise at first, but as more of the ice dissolves, the temperature falls.

To learn more about Sulfuric acid, refer to:

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8 0

1 year ago

If the volume of oxygen gas is 259 cm3 at 112 kPa, what would the volume be at 101.3 kPa?

NARA [144]

Answer:

<h3>The answer is 286.36 cm³</h3>

Explanation:

In order to find the volume be at 101.3 kPa we use Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question

P1 = 112 kPa = 112000 Pa

V1 = 259 cm³

P2 = 101.3 kPa = 101300 Pa

We have

$V_2 = \frac{112000 \times 259}{101300} = \frac{29008000}{101300} \\ = 286.3573543...$

We have the final answer as

Hope this helps you

6 0

3 years ago

50.0g of N2O4 is introduced into an evacuated 2.00L vessel and allowed to come to equilibrium with its decomposition product,N2O

Lady_Fox [76]

The mass of N2O4 in the final equilibrium mixture is 39.45 grams.

The decomposition reaction of N2O4 to 2NO2 can be expressed as:

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

From the parameters given:

The mass of N2O4 = 50.0 g
The molar mass of N2O4 = 92.011 g/mol

The number of mole of N2O4 can be determined as:

$\mathbf{Moles \ of \ N_2O_4 = \dfrac{50.0 g}{92.011 g/mol}}$

$\mathbf{Moles \ of \ N_2O_4 = 0.5434 \ moles}$

The volume of the vessel in which N2O4 was evacuated is = 2.0 L

From stochiometry, the concentration of $\mathbf{[N2O4] = \dfrac{mole \ of \ N_2O_4}{volume \ of \ N_2O_4}}$

$\mathbf{[N2O4] = \dfrac{0.5434}{2}}$

$\mathbf{[N2O4] = 0.2717 \ M}$

The I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.2717 0

Change -x +2x

Equilibrium (0.2717 - x) 2x

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(2x)^2}{(0.2717-x)}}$

Recall that; Kc = 0.133

∴

$\mathbf{0.133 = \dfrac{4x^2}{(0.2717-x)}}$

0.0361 - 0.133x = 4x²

4x² + 0.133x - 0.0361 = 0

By solving the above quadratic equation, we have;

x = 0.07978

The Concentration of [NO2] = 2x

[NO2] = 2 (0.07978)
[NO2] = 0.15956 M

The Concentration of [N2O4] = 0.2717 - x

[N2O4] = 0.2717 - 0.07978
[NO2] = 0.19192 M

Again, from the decomposition reaction, we can assert that;

$\mathbf{N_2O_4{(g)} \leftrightarrow 2NO_{2(g)}}$

0.19192 M of N2O4 decompose to produce 0.15956 M of NO2.

However, if 5.0 g of NO2 is injected into the vessel, then the number of moles of NO2 injected becomes;

$\mathbf{= \dfrac{5.0 \ g}{46 \ g/mol}} \\ \\ \\ \mathbf{= 0.10869\ moles}$

The Molarity of NO2 injected now becomes:

$\mathbf{= \dfrac{00.10869 }{2} } \\ \\ \\ \mathbf{= 0.05434 \ M }$

So, the new moles of [NO2] becomes = 0.15956 + 0.05434

= 0.2139 M

The new I.C.E table can be computed as:

$\mathbf{N_2O_4{(g)}}$ ↔ $\mathbf{ 2NO_{2(g)}}$

Initial 0.19192 0.2139 M

Change +x -2x

Equilibrium (0.19192 + x) (0.2139 -2x)

NOTE: The injection of NO2 makes the reaction proceed in the backward direction.

The equilibrium constant from the I.C.E table can be expressed as:

$\mathbf{K_c = \dfrac{[NO]^2}{[N_2O_4]}}$

$\mathbf{K_c = \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

Recall that; Kc = 0.133

$\mathbf{0.133= \dfrac{(0.2139 - 2x)^2}{(0.19192+x)}}$

By solving for x;

x = 0.2246 or x = 0.0225

We need to consider the value of x that is less than the initial concentration of NO2(0.2139 M) which is:

x = 0.0225

Now, the final concentration of [N2O4] = (0.19192 + 0.0225)M

= 0.21442 M

The final number of moles of N2O4 = Molarity(concentration) × volume

The final number of moles of N2O4 = (0.21442 × 2) moles

The final number of moles of N2O4 = 0.42884 moles

The mass of N2O4 in the final equilibrium mixture is:

= final number of moles × molar mass of N2O4

= 0.42884 moles × 92 g/mol

= 39.45 grams

Learn more about the decomposition of N2O4 here:

brainly.com/question/25025725

6 0

3 years ago

Calculate the mass of magnesium oxide produced when 0.645 g of magnesium hydroxide is the is completely decomposed.

FrozenT [24]

Answer:

0.44g

Explanation:

Given parameters:

Mass of magnesium hydroxide = 0.645g

Solution

The balanced reaction equation is shown below:

Mg(OH)₂ → MgO + H₂O

From the equation above, we see that the number of atoms are conserved.

To find the mass of magnesium oxide produced, we find the number moles of the decomposed magnesium hydroxide first:

Molar mass of Mg(OH)₂ = 24 + 2(16+1) = 58g/mol

Number of moles of Mg(OH)₂ = $\frac{mass}{molar mass}$

= $\frac{0.645}{58}$

= 0.011mole

Now, from the balanced equaiton, we know that:

1 mole of Mg(OH)₂ produces 1 mole of MgO

Therefore, 0.011mole of Mg(OH)₂ will produce 0.011mole of MgO

so mass of MgO = number of moles of MgO x molar mass of MgO

Molar mass of MgO = 24 + 16 = 40g/mol

mass of MgO = 0.011 x 40 = 0.44g

8 0

4 years ago