When Gas Becomes a Liquid.
When Gas Becomes a Solid.
When Liquid Becomes a Solid.
(Works for Plato Users) .
Mixture because your pouring something into it then stirring it i don’t know.
Explanation:
Carbon black sphere
Nitrogen blue sphere
Chlorine green sphere
Hydrogen white sphere
Reactants:
three molecules that contain one black sphere and four white spheres
3CH₄
three molecules that contain two blue spheres and four green spheres:
3N₂Cl₄
Products:
three molecules that contain one black sphere and four green spheres:
3CCl₄
three molecules that contain two blue spheres:
3N₂
six molecules that contain two white spheres
6H₂
The reaction proper:
3CH₄ + 3N₂Cl₄ → 3CCl₄ + 3N₂ + 6H₂
Learn more:
Chemical reactions brainly.com/question/4216541
#learnwithBrainly
<span>Answer: FALSE:
Explanation:
A reaction requires 22.4l of at STP. You have 32.0l of gas at 398k and 105.6 kpa.
1) STP stands for standard temperature and pressure.
2) Standard temperature is 0°C or 273.15 K
3) Standard pressure is 1 atm or 1013.25 kPa
4) use the ideal gas equation for both contidions
pV = n RT
=> n * R = pV /T
at STP n * R = 1031.25 kPa * 22.4 liter / 273.15 K = 84.5
at T = 398 K, p = 105.6 kPa, and V = 3.2.0 liter:
n * R = 105.6 kPa * 32.0 liter / 398 K = 8.49
Since R is a constant (the Universal Gases Constant), it is evident that the number of moles in the 32.0 liter of gas, at T = 398 K and P = 105.6 kPa is less than the number of moles of the 22.4 liter gas at STP.
There is not enough gas to carry out the reaction.
</span>
Answer:
See explanation
Explanation:
In this question, we have to remember that in all combustion reactions we will have <u>Oxygen</u> as reactive (
) and the products are <u>Carbon dioxide</u> (
) and <u>Water</u> (
). Additionally, for the states, we will have (l) for the <u>liquid state</u> and (g) for the <u>gas state</u>. So, we can analyze each reaction:
<u>Part A.</u>
The formula for nonane is 
, with this in mind we can write the combustion reaction:
When we balance the reaction we will obtain:
<u>Part B.</u>
The formula for 2-methylbutane is 
, with this in mind we can write the combustion reaction:
When we balance the reaction we will obtain:
<u>Part C.</u>
The formula for 3-ethyltoluene is 
, with this in mind we can write the combustion reaction:
When we balance the reaction we will obtain:
See figure 1 for further explanations.
I hope it helps!
