Answer:
compounds are elements include an element and a compound
Explanation:
elements in the decomposition reaction is the substance that cannot be separated into simpler substances. Compounds, technically act as a reactant in the decomposition reaction, but since the reaction breakdown one substance into two or more, sometimes it exists in the product
A. It does not retain the properties of the substances that make it up
Don't blame me if you get it wrong i'm a dumb but yet successful student
Warm blood gives deep-sea fish a boost, according to Wegner. The opah's muscles and nervous system likely function faster than an equivalent fish with cold blood. ... This fish, the southern opah, lives in colder waters than the northern opah, so it would be harder to keep warm, Wegner said — but even more beneficial.
The equilibrium constant of the reaction is 282. Option D
<h3>What is equilibrium constant?</h3>
The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.
Concentration of bromine = 0.600 mol /1.000-L = 0.600 M
Concentration of iodine = 1.600 mol/1.000-L = 1.600M
In this case, we must set up the ICE table as shown;
Br2(g) + I2(g) ↔ 2IBr(g)
I 0.6 1.6 0
C -x -x +2x
E 0.6 - x 1.6 - x 1.190
If 2x = 1.190
x = 1.190/2
x = 0.595
The concentrations at equilibrium are;
[Br2] = 0.6 - 0.595 = 0.005
[I2] = 1.6 - 0.595 = 1.005
Hence;
Kc = [IBr]^2/[Br2] [I2]
Kc = ( 1.190)^2/(0.005) (1.005)
Kc = 282
Learn more about equilibrium constant:brainly.com/question/15118952
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The balanced chemical reaction is written as:
<span>3NO2 + H2O = 2HNO3 + NO
Assuming that the gases in this reaction are ideal gas, then we can use the conversion from L to moles which is 1 mol of ideal gas is equal to 22.4 L. We calculate as follows:
538 L NO2 ( 1 mol / 22.4L ) ( 1 mol NO / 3 mol NO2 ) ( 22.4 L / 1 mol ) = 179.33 L NO is produced</span>
