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Citrus2011 [14]
2 years ago
6

can anyone draw out the answer for me please because I'm confused your supposed to make a lewis dot structure out of the lines a

nd dots showing electron transfer to create that iconic bond​

Chemistry
1 answer:
Andrews [41]2 years ago
8 0

Answer: are you allowed to type in letters?

An example of an ionic bond would be Li+ f- -> LiF or Ca+2 + s-2 -> CaS

Remember that ionic bonding is all about there charges

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i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i
9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

C_8H_{18}+O_2\to CO_2+H_2O

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

C_8H_{18}+O_2\to8CO_2+H_2O

Now, we balance the hydrogen:

C_8H_{18}+O_2\to8CO_2+9H_2O

And in the end, the oxygen:

C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O

We can multiply all coefficients by 2 to get integer ones:

2C_8H_{18}+25O_2\to16CO_2+18H_2O

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}

For the products, we have:

\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}

Now, we substract the rectants from the produtcs:

\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}

Then, we need to melt all this mass, so we use the latent heat now:

Q_2=n\cdot6.03kJ/mol

Converting mass to number of moles of water we have:

\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}

So:

Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g

Adding them, we have a total heat of:

\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}

Since we have a total of 20kg of ice, we can clculate the percent using it:

P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%

5 0
11 months ago
3) _Na +<br> |H,80_ - _Na,304<br> _Na2SO4 + _H2
erik [133]
Hey


How was your day
3 0
2 years ago
What is the difference between electron behavior in ionic and covalent bonds?
Flauer [41]

a electron behavior is a subatomic particle whose electric i charge and the other one is a molecule and chemical bond that shares electrons

8 0
3 years ago
If the temperature on 244 mL of a gas is changed to 488 mL and 6 atm, at constant
Fittoniya [83]

Answer:

<h2>12 atm</h2>

Explanation:

To find the initial pressure we use the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the initial pressure

P_1 =  \frac{P_2V_2}{V_1}  \\

From the question we have

P_1 =  \frac{488 \times 6}{244}  =  \frac{2928}{244}  \\

We have the final answer as

<h3>12 atm</h3>

Hope this helps you

6 0
2 years ago
SO2(g) + NO2(g) ↔ SO3(g) + NO(g) Kc = 0.33 A reaction mixture contains 0.41 M SO2, 0.14 M NO2, 0.12 M SO3 and 0.14 M NO. Which o
stepladder [879]

Answer:

The reaction will shift in the direction of products.

Explanation:

<u>Step 1:</u> Data given

A reaction mixture contains:

0.41 M SO2

0.14 M NO2

0.12 M SO3

0.14 M NO

<u>Step 2:</u> The balanced equation

O2(g) + NO2(g) ↔ SO3(g) + NO(g)      Kc = 0.33

<u>Step 3:</u> Define the direction of the shift of  reaction:

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

<u>Step 4:</u> Calculate Q

Q = [NO][SO3]/[SO2][NO2]

Q = (0.14 *0.12)/(0.41*0.14)

Q = 0.0168/0.0574

Q = 0.293

Q<Kc

This means there are more reactants than products. Thud, some of the reactants will become products, causing the reaction to shift to the right.

The reaction will shift in the direction of products.

8 0
3 years ago
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