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Ray Of Light [21]
2 years ago
9

To burn 1 molecule of C5H12 to form CO2 and H2O (complete combustion), how many molecules of O2 are required?

Chemistry
2 answers:
viva [34]2 years ago
5 0
C5H12 + 8 O2 → 5 CO2 + 6 H2O 

8 molecules of O2 are required.
matrenka [14]2 years ago
5 0

Answer:

8 molecules of oxygen gas are required.

Explanation:

Combustion reaction is the type of chemical reaction in which a hydrocarbon reacts with oxygen gas to give to carbon dioxide and water.

C_5H_{12}+8O_2\rightarrow 5CO_2+6H_2O

According to reaction ,1 mole of pentane reacts with 8 molecules of oxygen gas to give 5 moles of carbon dioxide and 6 moles of water.

So, to burn 1 mole of pentane we will need 8 molecules of oxygen gas.

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What is the best description of what has happened to flowers when they change color as they mature?
SVEN [57.7K]

Answer:

The answer is (A)

Explanation:

When the weather changes, nature also changes because most plants rely on photosynthesis and if they don't get as much light then they can't support as much as they used causing them to shut down parts of the plant.

6 0
2 years ago
How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?
Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

Molality=\frac{n\times 1000}{W_s}

where,

n= moles of solute

Moles of NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles  

W_s = weight of the solvent in g = ?

0.452=\frac{0.348\times 1000}{W_s}

W_s=770g

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

5 0
2 years ago
What is the molar mass of 56.75 g of gas exerting a pressure of 2.87 atm on the walls of a 5.29 l container at 230 k?
laiz [17]
We first need to find the number of moles of gas in the container 
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation 
290 802.75 Pa x  5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol 
therefore molar mass is 70.6 g/mol 
8 0
2 years ago
USATEST PREP QUESTION HELP ! Really appreciate it !
Novosadov [1.4K]

Answer is A Access pictures of the area taken by satelites.

Explanation: Satelites are the only thing out of these four answers that does not requir power supply from the town. Hope it helped!

5 0
2 years ago
Read 2 more answers
What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.
Mamont248 [21]

Answer:

78.46 grams of  2-bromopropane could be prepared from 25.5 g of propene

Explanation:

C_3H_6+HBr\rightarrow C_3H_7Br

Moles of propene = \frac{25.5 g}{39 g/mol}=0.6538 mol

According to reaction, 1 mole of propene gives 1 mole of propane.

Then 0.6538 moles of bromo-propane will give:

0.6538 mol\times 120 g/mol=78.46 g

78.46 grams of  2-bromopropane could be prepared from 25.5 g of propene.

5 0
2 years ago
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