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anzhelika [568]
3 years ago
10

Ethane (C2H6) is one of many compounds that burns in oxygen to form carbon dioxide and water, as represented by the equation bel

ow. Which substance is a product in this reaction? *
Chemistry
1 answer:
Ulleksa [173]3 years ago
5 0

Answer: From the question, the equation which is 2C2H6 + 7O2 --> 4CO2 + 6H2O has the products as 4CO2 + 6H2O( which is carbon dioxide and water)

Explanation:

Chemical reaction is a process in chemistry which leads to the formation of new substances referred to as PRODUCTS from combination of two or more substances called REACTANTS. There are different types of chemical reaction which includes COMBUSTION REACTION that is mostly seen in hydrocarbons.

Ethane is an example of a saturated hydrocarbon ( that is all the four bonds of carbon are single bonds) which undergoes combustion reaction with oxygen molecule. Ethane will burn in air( oxygen) to give off carbon dioxide and water as products of the chemical reaction.

2C2H6 + 7O2 --> 4CO2 + 6H2O

Shane and oxygen is the reactants of the chemical reaction while carbon dioxide and water is the products.

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What would be the minimum energy Emin required to excite a hydrogen atom from its lowest energy level
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3 0
2 years ago
What is the concentration if 20 grams of NaOH was dissolved in enough water to make a 250 mL NaOH solution?
gizmo_the_mogwai [7]

Answer: 1 molar NaOH contains 40 grams of NaOH per every liter of water

Explanation:

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3 0
3 years ago
On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

6 0
3 years ago
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