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bazaltina [42]
3 years ago
5

The total energy radiated by a blackbody depends on.

Chemistry
1 answer:
polet [3.4K]3 years ago
5 0
Absolute temperature.
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When dissolving a solid or liquid, as the temperature of the solvent increases, the rate of dissolution?
Alinara [238K]

Answer:A. Increases

Explanation:

Heating or an increase in temperature increases the kinetic energy of particles thereby increasing their motion and how they relate and react with one another.

Increase in the temperature of the solvent is directly proportional to the rate of dissolution. The rate of dissolution increases due to the increase in kinetic energy. This makes the solvent particles interact faster with the solute particles thereby increasing the dissolution rate.

7 0
3 years ago
A compound x is an important ingredient of an antacid. It is also used in fine extinguishers. Identify x.
Maslowich
NaHCO3 is the required substance often used in antacid and is used in Soda Acid Fire Extinguisher.
3 0
3 years ago
Read 2 more answers
Describe the relationship between an enzyme, substrate, and active site.
alexdok [17]
So what I know is that enzyme and substrate are like lock and key meaning that when the active site of the enzyme changes, the enzyme will not fit to the substrate which will lead the enzyme to denature. Hope this helps.
8 0
3 years ago
A solution of water (kf=1.86 ∘c/m) and glucose freezes at − 2.15 ∘c. what is the molal concentration of glucose in this solution
Reil [10]
<span>1.16 moles/liter The equation for freezing point depression in an ideal solution is ΔTF = KF * b * i where ΔTF = depression in freezing point, defined as TF (pure) â’ TF (solution). So in this case ΔTF = 2.15 KF = cryoscopic constant of the solvent (given as 1.86 âc/m) b = molality of solute i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1. Solving for b, we get ΔTF = KF * b * i ΔTF/KF = b * i ΔTF/(KF*i) = b And substuting known values. ΔTF/(KF*i) = b 2.15âc/(1.86âc/m * 1) = b 2.15/(1.86 1/m) = b 1.155913978 m = b So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
7 0
3 years ago
According to the Arrhenius concept, which of the following substances is NOT a base in aqueous solutionsA. NH3B. LiOHC. NaOHD. H
olga_2 [115]

Answer:

NH₃ (Option A)

Explanation:

Arrhenius theory explained that the acids are the ones that have H⁺, either H in its formula. Following this, the bases are the ones that have OH⁻ , either OH and its formula.

It can be used only with compounds with H, or OH.

So the ammonia is not a base, as Arrhenius theory.

It is known that ammonia behaves as a weak base, but it does not have hydroxide ions that can yield to water

5 0
3 years ago
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