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yulyashka [42]
3 years ago
10

The products of a combustion reaction do NOT include ____. (1 point) water carbon dioxide carbon monoxide hydrogen The products

of a combustion reaction do NOT include Hydrogen.
Chemistry
2 answers:
nalin [4]3 years ago
7 0

the answer is water

Hope this help1

Allushta [10]3 years ago
6 0

Answer: Option (d) is the correct answer.

Explanation:

In combustion reaction, a compound reacts with oxygen and results in the formation of carbondioxide and water.

For example, when propane reacts with oxygen it results in the formation of carbondioxide and water.

The reaction is as follows.

      C_{3}H_{8} + O_{2} \rightarrow CO_{2} + H_{2}O

Also, when there is incomplete combustion of carbon then it results in the formation of carbon monoxide. This reaction is as follows.

         C + \frac{1}{2}O_{2} \rightarrow CO

Thus, we can conclude that the products of a combustion reaction do NOT include hydrogen.

You might be interested in
Analyze the food web. If the frog population decreases, what impact will this likely have on the food web?
sergiy2304 [10]

Answer:

the population of grasshoppers will decrease

Explanation:

a food source is becoming more and more limited thus, its predator will also start to die out due to starvation

8 0
3 years ago
Read 2 more answers
A balloon contains 0.140 molmol of gas and has a volume of 2.78 LL . If an additional 0.152 molmol of gas is added to the balloo
levacccp [35]

Answer:

The final volume will be 5.80 L

Explanation:

Step 1: Data given

Number of moles gas = 0.140 moles

Volume of gas = 2.78 L

Number of moles added = 0.152 moles

Step 2: Calculate the final volume

V1/n1 = V2/n2

⇒ with V1 = the initial volume = 2.78 L

⇒ with n1 = the initial number of moles = 0.140 moles

⇒ with  V2 = The new volume = TO BE DETERMINED

⇒ with n2 = the new number of moles = 0.140 + 0.152 = 0.292 moles

2.78/0.140 = V2 /0.292

V2 = 5.80 L

The final volume will be 5.80 L

8 0
3 years ago
Can anyone help me on this please?
Bad White [126]
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one.  You can't
bluff through it.


7 0
3 years ago
If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give
tankabanditka [31]

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Given:

The nitrogen gas molecule with a temperature of 330 Kelvins is released from Earth's surface to travel upward.

To find:

The maximum height of a nitrogen molecule when released from the Earth's surface before coming to rest.

Solution:

  • The maximum height attained by nitrogen gas molecule = h
  • The temperature of nitrogen gas particle = T = 330 K

The average kinetic energy of the gas particles is given by:

K.E=\frac{3}{2}K_bT\\\\K.E=\frac{3}{2}\times 1.38\times 10^{-23} J/K\times 330 K\\\\K.E=6.381\times 10^{-21} J

The nitrogen molecule at its maximum height will have zero kinetic energy as all the kinetic energy will get converted into potential energy

  • The potential energy at height h = P.E = 6.381\times 10^{-21} J
  • Molar mass of nitrogen gas =  28.0134 g/mol
  • Mass of nitrogen gas molecule = m

m= \frac{ 28.0134 g/mol}{6.022\times 10^{23} mol^{-1}}=4.652\times 10^{-23} g\\\\1g=0.001kg\\\\m=4.652\times 10^{-23}\times 0.001 kg\\\\=4.652\times 10^{-26} kg

  • The acceleration due to gravity = g = 9.8 m/s^2
  • The maximum height attained by nitrogen gas molecule = h
  • The potential energy is given by:

P.E=mgh

6.381\times 10^{-21} J=4.652\times 10^{-26} kg\times 9.8 m/s^2\times h\\\\h=\frac{6.381\times 10^{-21} J}{4.652\times 10^{-26} kg\times 9.8 m/s^2}\\\\h=13,996.6 m\\\\1 m = 0.001 km\\\\h=13,996.6 m=h=13,996.6\times 0.001 k m\\\\=13.9966 km \approx 14 km

The maximum height at which nitrogen molecule will go before coming to rest is 14 kilometers.

Learn more about the average kinetic energy of gas particles here:

brainly.com/question/16615446?referrer=searchResults

brainly.com/question/6329137?referrer=searchResults

6 0
2 years ago
A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c
makkiz [27]
0.216g of aluminium compound X  react with an excess of water water to produce gas. this gas burn completely  in O2  to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n  =  24 / 1 
n = 0.0045 mole ( CO2 >>0.0045 mole 
0.216 - 0.0045 = 0.2115
so Al =   0.2115 / 27  =>  0.0078 mole 
C = 0.0045 * 1000 => 4.5    and Al  = 0.0078 * 1000 = 7.8 

7 0
3 years ago
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