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dezoksy [38]
3 years ago
8

Given the incomplete equation for the combustion of ethane:

Chemistry
2 answers:
andriy [413]3 years ago
5 0

Answer : The correct option is, (4) (H_2O)

Explanation :

Combustion reaction : It is a reaction in which a hydrocarbon react with the oxygen gas to produces carbon dioxide as water as a products.

The given incomplete equation are :

2C_2H_6+7O_2\rightarrow 4CO_2+6_

In this equation, we see that a hydrocarbon and oxygen gas are present on reactant side and carbon dioxide are present on product side but water molecules are missing on the product side. So, we are adding water (H_2O) on missing side.

The given complete equation will be :

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

Therefore, the formula missing on product is,  (H_2O)

Svetach [21]3 years ago
3 0
The combustion of ethane will consume oxygen and create carbon dioxide and water. That is a general rule for most alkane, olefin and alkyne. So the missing product is H2O.
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Under the same conditions of temperature and pressure a liquid differs from a gas because the particles of the liquid
noname [10]

Answer:

liquid has more attraction between molecules. It takes energy to break these forces to change the liquid to a gas.

Explanation:

because liquid has more attraction

6 0
3 years ago
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In a sulphuric acid (h2so4) - sodium hydroxide (naoh) acid-base titration, 17.3 ml of 0.126 m naoh is needed to neutralize 25 ml
katen-ka-za [31]
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted  - 0.126 mol/L  x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄ 
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄ 
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol 
then 1000 mL contains - 0.00109 mol / 25 mL  x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
4 0
3 years ago
What is the mass of volume= 5.4 mL  density= 2.5 g/mL​
ddd [48]

Answer:

13.5g

Explanation:

Mass is defined as the measure of the amount of matter in an object. Its unit is kg or g.

Mass can be calculated using the formula:

m= d × v

where,

d= density

m= mass

v= volume

m= 2.5×5.4

m= 13.5g

5 0
3 years ago
Identify the independent variable, dependent variable, and constant in the following problem: If you plant hydrangeas in your ga
Afina-wow [57]
The independent variable is the one you change, and the dependent variable is the one that changes because the independent changed (i think). The constant variable is the thing you can't change so the experiment is fair.

Independant: whatever causes the flowers to change color 
Dependent: the flowers change color
Constant: same soil type, same conditions, etc.
5 0
3 years ago
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Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.
Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

\%C_E = \frac{E}{T} * 100\%

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In (Ca(OH)_2)

The amount of all elements is: (i.e formula mass of (Ca(OH)_2))

T = 1 * Ca + 2 * H + 2 * O

T = 1 * 40 + 2 * 1 + 2 * 16

T = 74

The amount of calcium is: (i.e formula mass of calcium)

E = 1 * Ca

E = 1 * 40

E = 40

So, the percentage component of calcium is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{40}{74} * 100\%

\%C_E = \frac{4000}{74}\%

\%C_E = 54.05\%

The amount of hydrogen is:

E = 2 * H

E = 2 * 1

E = 2

So, the percentage component of hydrogen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{2}{74} * 100\%

\%C_E = \frac{200}{74}\%

\%C_E = 2.70\%

Similarly, for oxygen:

The amount of oxygen is:

E = 2 * O

E = 2 * 16

E = 32

So, the percentage component of oxygen is:

\%C_E = \frac{E}{T} * 100\%

\%C_E = \frac{32}{74} * 100\%

\%C_E = \frac{3200}{74}\%

\%C_E = 43.24\%

5 0
2 years ago
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