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Makovka662 [10]
2 years ago
7

a burning candle wax reacts with oxygen. after a glass jar is placed over it, the candle goes out. what is the limiting reactant

?
Chemistry
2 answers:
trapecia [35]2 years ago
5 0

Answer:

The Oxygen

Explanation:

When fire is covered, it goes out due to the lack of oxygen to maintain itself.

USPshnik [31]2 years ago
4 0

The limiting reactant in the experiment of the burning candle is oxygen.

<h3>What is a limiting reactant?</h3>

A limiting reactant is a reactant which determines the amount of product that ca be formed in a given reaction.

The limiting reactant is always used up in a reaction, after which the reaction stops.

In the experiment of the burning candle, covering the candle with the glass prevents more oxygen from taking part in the reaction.

Therefore, the limiting reactant is oxygen.

Learn more about limiting reactant at: brainly.com/question/14225536

#SPJ11

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After mg ribbon is reacted completely with air, what is the reason to add some water and heat?
Vikentia [17]

The reason why some water and heat are added after Mg ribbon is reacted completely with air is to remove Mg3N2 as MgO.

Combustion reaction is a reaction in which a substance is completely burnt in oxygen. Magnesium ribbon can be converted to magnesium oxide by combustion in air.

The reaction is as follows; 2Mg(s) + O2 ----> 2MgO(s)

Since air contains nitrogen, Mg3N2 is also formed. Heat and water can be added to the reaction thereby converting magnesium nitride to magnesium oxide.

Learn more: brainly.com/question/11527546

3 0
2 years ago
Integrated rate law for second order unimolecular irreversible
kirill115 [55]

Answer:

The rate law for second order unimolecular irreversible reaction is

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}

rearranging the ecuation

-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}

Integrating between times 0 to <em>t </em>and between the concentrations of [A]_{0} to <em>[A].</em>

\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}

Solving the integral

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

5 0
3 years ago
Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc
marysya [2.9K]

Explanation:

1) 2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)

Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091

2) H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)

Kc=\frac{[H_2SO_4]}{[SO_3]^2}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3

3)  P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)

Kc=\frac{1}{[O_2]^3}

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641

5 0
3 years ago
When an irregularly shaped chunk of an unknown metal with a mass of 25.32 g was placed in a graduated cylinder containing 25.00
inn [45]

Answer: The density of the unknown metal is 7.86 g/ml.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of metal  = 25.32 g

Volume of metal = volume of water displaced = (28.22 - 25.00) ml = 3.22 ml

Putting in the values we get:

density=\frac{25.32g}{3.22ml}

Density=7.86g/ml

Thus the density of the unknown metal is 7.86 g/ml

7 0
3 years ago
What colour is litmus in a solution of pH 4?
schepotkina [342]

Answer:

in solution of PH4 it is red

in water it is blue

6 0
3 years ago
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