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3 years ago

Q82. Liquid petrol does not ignite spontaneously when exposed to the air because

1 answer:

7 0

Liquid petrol does not ignite spontaneously when exposed to the air because there is not enough molecules that contains energy to reach the activation energy. As a result, the reaction cannot proceed and a supply of energy should be added for the reaction to occur.

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Fill in the blanks with the word bank. !!URGENT NEED TODAY!!

stiv31 [10]

blank 1: lost

blank 2: gained

blank 3: same

blank 4: side

blank 5: equation

blank 6: coefficients

blank 7: formulas

blank 8: change

6 0

3 years ago

A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix

Alisiya [41]

Answer:

$M_f=38.8\%$

Explanation:

From the question we are told that:

Pressure $P=747mmHg$

Temperature $T=298K$

Volume $V=11.1$

Heat Produced $Q=780kJ$

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

$n= (747/760) *11.1/ (0.0821*298)$

$n=0.446mol$

Therefore

$x+y=0.446$

$x=0.446-y .....1$

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

$x(889) + y(2220) = 760 ...... 2$

Comparing Equation 1 and 2 and solving simultaneously

$x=0.446-y .....1$

$x(889) + y(2220) = 760 ...... 2$

$x=0.173$

$y=0.273$

Therefore

Mole fraction 0f Methane is mathematically given as

$M_f=\frac{x}{n}*100\%$

$M_f=\frac{1.173}{0.446}*100\%$

$M_f=38.8\%$

7 0

3 years ago

Timmy and his dad are throwing rocks into the pond. The rocks that Timmy's dad throws go farther and faster. Which of

7nadin3 [17]

Can you put a picture

3 0

2 years ago

Cell and chromosome characteristics?

rusak2 [61]

Answer:

hope the inserted image will help :)

Explanation:

5 0

3 years ago

Read 2 more answers

Consider the following intermediate chemical equations.

QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$ $\Delta H_1=-2439kJ$ (1)