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VikaD [51]
2 years ago
6

What is freshly pumped oil called?

Chemistry
2 answers:
algol132 years ago
7 0

Answer:

When the drill hits oil, some of the oil rises from the ground high into the air. This immediate release of oil is known as a "gusher." Once a reservoir has been located, pumps are used to extract the oil.

ser-zykov [4K]2 years ago
6 0

What is freshly pumped oil called?

petroleum

Explanation:

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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
1 year ago
how much heat, in terms in q, would it take to produce the change in temperature indicated in the picture? what is your reasonin
STALIN [3.7K]

Answer:

1. q.

2. 2q.

3. 3q.

4. 6q.

Explanation:

We'll begin by calculating the specific heat capacity of the liquid. This can be obtained as follow:

Mass (m) = 25 g

Change in temperature (ΔT) = 20 °C

Heat (Q) = q

Specific heat capacity (C) =?

Q = MCΔT

q = 25 × C × 20

q = 500C

Divide both side by 500

C = q/500

C = 2×10¯³ qg°C

Therefore, the specific heat capacity of liquid is 2×10¯³ qg°C

Now, we shall determine the heat required to produce the various change in temperature as follow:

2. Mass (m) = 50 g

Change in temperature (ΔT) = 20 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 20

Q = 2q.

Therefore, the heat required is 2q.

3. Mass (m) = 25 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 25 × 2×10¯³ × 60

Q = 3q.

Therefore, the heat required is 3q.

4. Mass (m) = 50 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 60

Q = 6q.

Therefore, the heat required is 6q.

4 0
3 years ago
Estimate the air pressure at an altitude of 5km
r-ruslan [8.4K]
The answer is 100 kg
6 0
3 years ago
Some elements are liquids or gases.<br><br><br> True <br> False
gregori [183]
The answer to this ? is true
4 0
3 years ago
A container of gas has a volume of 280 mL at a temperature of 22 Celsius if the pressure remains constant what is the volume 44
d1i1m1o1n [39]

Answer:

300.9mL

Explanation:

Given parameters:

V₁  = 280mL

T₁ = 22°C

T₂  = 44°C

Unknown:

V₂ = ?

Solution:

To solve this problem, we apply Charles's law;

  it is mathematically expressed as;

            \frac{V_{1} }{T_{1} }  = \frac{V_{2} }{T_{2} }

We need to convert the temperature to kelvin;

       T₁ = 22°C = 22 + 273 = 295K

       T₂  = 44°C = 44 + 273  = 317K

Input the parameters and solve;

        \frac{280}{295}   = \frac{V_{2} }{317}

  V₂ x 295 = 280 x 317

  V₂ = 300.9mL

5 0
3 years ago
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