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Vikki [24]
3 years ago
6

Give the balanced chemical equation (including phases) that describes the combustion of butene, C4H8(g). Indicate the phases usi

ng abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Express your answer as a chemical equation.
Chemistry
2 answers:
Studentka2010 [4]3 years ago
7 0
Butene is an alkene, which is a hydrocarbon that contains a double bond. At room temperature, butene occurs as a gas. So, when it reacts with oxygen gas, it will yield gaseous products, carbon dioxide and water vapor. The complete balance reaction is written below:

C₄H₈ (g) + 6 O₂ (g) → 4 CO₂ (g) + 4 H₂O (g)
Naddik [55]3 years ago
5 0

Answer:

The chemical equation is given as:

C_4H_8+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

Explanation:

Combustion reaction is the type of chemical reaction in which an organic compound undergoes reaction with oxygen gas to give carbon dioxide and water product.

When butene undergoes combustion reaction it gives carbon dioxide gas and water vapors as a product.

The chemical equation is given as:

C_4H_8+6O_2(g)\rightarrow 4CO_2(g)+4H_2O(g)

According to stoichiometry, 1 mole of butene gas reacts with 6 moles of oxygen gas to give 4 moles of carbon dioxide and 4 moles water vapors.

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The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins
saul85 [17]
The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>t_{1/2} = 2.5 min

We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{10 min}{2.5 min}
⇒ n=4<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>(1/2)^{n} = x
</span>⇒ x=(1/2)^4
<span>⇒x= \frac{1}{16}</span>
3 0
3 years ago
The table describes how some substances were formed.
lisabon 2012 [21]

Explanation:

Which is a pure substance?

1. soda

2. gasoline

3. salt water

4. carbon dioxide

carbon dioxide

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of -7.2°C. Bromine can be classified as a

1. compound.

2. impure substance.

3. mixture.

4. pure substance.

pure substance.

6 0
3 years ago
Read 2 more answers
A 118-ml flask is evacuated and found to have a mass of 97.129 g. when the flask is filled with 768 torr of helium gas at 35 ?c,
Inessa05 [86]
The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.

To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.

From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.

1) From pV = nRT, n = pV / RT

Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K

n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol  * 308.15K] =0.00472 mol

mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042

=> MM =  mass/n = 0.042 / 0.00472 = 8.90 g/mol

Now from a periodic table or a table you get that the molar mass of He is 4g/mol

So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
7 0
3 years ago
which pair shares the same empirical formula. c2h2 and c6h6, c2h2 and c2h4, ch2 and c6h6, ch and c2h4
olasank [31]

Answer:

Answer: CH₃ and C₂H₆ have same empirical formula.

Explanation:

it just compares in that it its the same

7 0
3 years ago
A gas bubble has a volume of 0.650 mL at the bottom of a lake, where the pressure is
astraxan [27]

Considering the Boyle's law, as the pressure decreases, volume increases and has a value of 2.246 mL.

<h3>Boyle's law</h3>

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

This law says that the volume occupied by a given mass of gas at constant temperature is inversely proportional to the pressure. This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

P×V=k

If an initial state 1 and a final state 2 are analyzed, Boyle's law is expressed as:

P1×V1=P2×V2

<h3>Volume at the surface of the lake</h3>

In this case, you know:

  • P1= 3.46 atm
  • V1= 0.650 mL
  • P2= 1 atm
  • V2= ?

Replacing in Boyle's law:

3.46 atm× 0.650 mL= 1 atm×V2

Solving:

V2= (3.46 atm× 0.650 mL)÷ 1 atm

<u><em>V2= 2.246 mL</em></u>

Finally, as the pressure decreases, volume increases and has a value of 2.246 mL.

Learn more about Boyle's law:

brainly.com/question/4147359

#SPJ1

4 0
2 years ago
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