Pls helps fast test due soon!!!

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

20 points - Chemistry - plzzz help

Angelina_Jolie [31]

Answer:

1. Inversely proportional

2. Option C. Boyle's Law

3. Directly proportional

4. Option C. Gay-Lussac's Law

5. Directly proportional

6. Option C. Charles' Law

Explanation

Boyle's law states that the volume of a fixed mass of gas is inversely proportional to the pressure provided temperature remains constant. Mathematically,

V & 1/P

V = K/P

PV = K(constant)

Charles' law states that the volume of a fixed mass of gas is directly proportional to it's absolute temperature, provided pressure remains constant. Mathematically,

V & T

V = KT

V / T = K(constant)

Gay-Lussac's Law states that the pressure of a fixed mass of gas is directly proportional to it's absolute temperature, provided the volume remains constant. Mathematically

P & T

P = KT

P/ T = K (constant)

7 0

3 years ago

If a decrease in temperature accompanies a reaction, what occurred?

Sliva [168]

Answer:

Energy was released

Explanation:

Decrease in temperature means less energy, but the energy had to have been transferred because energy can't be destroyed or created. So if the temp. went down, energy must have been lost. (If it had been absorbed, the temp. would have gone up)

3 0

2 years ago

Read 2 more answers

5. __NH3 + __O2 >>>>>__ NO +__ H20

inn [45]

2 NH3+ 2 O2 —> 2 NO+ 3 H2O

5 0

3 years ago

What group is helium in? Group 8 or group 2?

Gemiola [76]

Answer:

group 8 of the periodic table