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Jet001 [13]
3 years ago
10

How to tell what side is the reactant and what side is the products?

Chemistry
1 answer:
Julli [10]3 years ago
4 0
Reactant at the left product at the rig
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If you ask for water and water immediately appears that very second, this is an example of something instantaneous.

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In a certain time, light travels 4.96 km in a vacuum. during the same time, light travels only 3.36 km in a liquid. what is the
Natali5045456 [20]

the refractive index of the liquid is 1.476

The refractive index, which has no dimensions, measures how quickly light passes through a substance.

It can also be described as the difference between the speed of light in a vacuum and a medium.

Refractive index is equal to the product of the light's liquid and vacuum speeds.

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speed of light in vacuum = 4.96 km/t

speed of light in liquid = 3.36 km/t

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Refractive index =1.476

Therefore, the refractive index of the liquid is 1.476

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1 year ago
How are carbon-based molecules suited for sustaining life?
IrinaVladis [17]

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Life on Earth is based on carbon, likely because each carbon atom can form bonds with up to four other atoms simultaneously. This quality makes carbon well-suited to form the long chains of molecules that serve as the basis for life as we know it, such as proteins and DNA.

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8 0
3 years ago
What occurs to the atoms of reactants in a chemical reaction?
djyliett [7]

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What occurs to the atoms of reactants in a chemical reaction?

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In a chemical reaction, the atoms and molecules that interact with each other are called reactants. ... No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

6 0
3 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
4 years ago
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