If acetone has a density of 0.7857
the mass in grams of point A is 22.4 g and the volume at point B is 8.32 mL.
<h3>What is acetone?</h3>
Acetone is known as a chemical substance that is usually found in the environment but can also be produced artificially. Acetone is a polar organic product that interacts very well with water molecules, generating dipole-dipole relationships.It is colorless with a distinctive smell and taste, we find it in products known as <u>cleaning and personal care products</u>, but we can also use it as a solvent for substances.
Also in the environment in <u>plants, trees and in volcano emissions or in forest fires</u>, it does not become <em>toxic</em> in low doses but if it is exposed to an individual in high doses it can become <em>fatal</em>.
In the statement we can find that acetone has a density of 0.7857
.
Therefore, we can confirm that if acetone has a density of 0.7857
the mass in grams of point A is 22.4 g and the volume at point B is 8.32 mL.
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Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
Hydroxylamine in water: HONH₂(aq) + H₂O(l) ⇄ HONH₃⁺(aq) + OH⁻(aq).
Hydroxylammonium nitrate in water: HONH₃NO₃(aq) → OHNH₃⁺(aq) + NO₃⁻(aq).
1) with positive hydrogen ions (protons) react base and gives weak conjugate acid:
H⁺(aq) + HONH₂(aq) ⇄ HONH₃⁺(aq).
2) with hydroxide anions react acid and produce weak base and weak electrolyte water:
HONH₃⁺(aq) + OH⁻(aq) ⇄ HONH₂(aq) + H₂O(l).
Answer:
i don't know myself either