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2 years ago

What will happen when a wax candle burned in air with limited oxygen?

Chemistry

2 answers:

garik1379 [7]2 years ago

7 0

Answer:

if the candle is burned in air with limited oxygen supply the candle will go out of fire as the oxygen is reducing....

Explanation:

Hope It Helps!!!

vfiekz [6]2 years ago

6 0

Answer:

The molecules of carbon dioxide are heavier than air. When a result, as they descend over the flame and candle, they push the oxygen and other air molecules out of the path. The oxygen can no longer react with the wax when it is pushed away from the wick. The flame is put out as a result.

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Carbon Monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K, the equilibrium constant is 5.10. Calculate t

ozzi

Answer: Concentration of $CO_2$ at equilibrium= 1.386 M

Concentration of $H_2$ at equilibrium = 1.386 M

Concentration of $CO$ at equilibrium = 0.614 M

Concentration of $H_2O$ at equilibrium= 0.614 M

Explanation:

Moles of $CO$ = 1.00 mole

Moles of $H_2O$ = 1.00 mole

Moles of $CO_2$ = 1.00 mole

Moles of $H_2O$ = 1.00 mole

Volume of solution = 1.00 L

Initial concentration of $CO$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $H_2O$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $CO_2$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

Initial concentration of $H_2$ = $\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M$

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2$

Initial conc. 1.00M 1.00 M 1.00 M 1.00 M

At eqm. conc. (1.00-x) M (1.00-x) M (1.00+x) M (1.00+x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}$

Now put all the given values in this expression, we get :

$5.10=\frac{(1.00+x)^2}{(1.00-x)^2}$

By solving the term 'x', we get :

x = 0.386

Concentration of $CO_2$ at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of $H_2$ = (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of $CO$ = (1.00-x) M = (1.00-0.386) M = 0.614 M

Concentration of $H_2O$ = (1.00-x) M = (1.00-0.386) M = 0.614 M

3 0

3 years ago

Calculate moles of naoh 38.18ml of 0.550M

USPshnik [31]

38.18ml to 0.550 you are going to have to divide that by the dividen of 550 ok then when you have done that msg me kk

3 0

3 years ago

when the volume of a gas is changed from 3.75 L to 6.52 L the temperature will change from 100k to _K

harkovskaia [24]

The temperature will change from 100K to 173.87 K

calculation

by use of law that is V1/T1=V2/T2

V1=3.75 L

T1=100k

V2=6.53 L

T2=?

make T2 the subject of the formula

T2=(V2 xT1)V1

=6.52 x100/3.75=173.87K

4 0

3 years ago

Read 2 more answers

A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, w

Fed [463]

Answer:

Density = 8.92 g/cm³

Explanation:

Density shows the relation of the mass and volume of a determined object. We have this deffinition:

Density = mass / volume

First of all, we calcualte the volume of the block of copper metal, with the data given:

8.4 cm . 5.5 cm . 4.6 cm = 212.52 cm³

Now we replace at the density formula:

Density = 1896 g /212.52 cm³ = 8.92 g/cm³

7 0

3 years ago

Which of the following is the correct representation of Gibbs’ Free Energy equation?

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5 0

3 years ago

Read 2 more answers

What will happen when a wax candle burned in air with limited oxygen?​

What will happen when a wax candle burned in air with limited oxygen?