Decomposition reaction D. 2H20⇒ 2H2 + O2
<h3>Further explanation
</h3>
Every chemical reaction involves a compound consisting of reactants and products
Reactants are compounds that react and form new compounds called products
There are several forms of reactions that can occur, including single replacement, double replacement, synthesis, decomposition, etc.
A. 2C2H6 + 702 ⇒ 4CO2 + 6H20
Combustion : reaction of Hydrocarbon and Oxygen
B. AgNO3 + LiCl → AgCl + LINO3
Double replacement : there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product
C. Ca + MgS → CaS + Mg
Single replacement :one element replaces the other elements of a compound to produce new elements and compounds
D. 2H20⇒ 2H2 + O2
Decomposition : One compound breaks down into 2 components
Answer:
648.5 mL
Explanation:
Here we will assume that the pressure of the gas is constant, since it is not given or specified.
Therefore, we can use Charle's law, which states that:
"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"
Mathematically:

where
V is the volume of the gas
T is its absolute temperature
The equation can be rewritten as

where in this problem we have:
is the initial volume of the gas
is the initial temperature
is the final temperature
Solving for V2, we find the final volume of the gas:

Answer:Lithium
Explanation:Lithium has 3 protons, and 4 neutrons making it the lightest element that isn't a gas.
Answer:

Explanation:
Hello there!
Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
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