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storchak [24]
2 years ago
12

The chemical formula for acetone is: ch32co calculate the molar mass of acetone. round your answer to 2 decimal places.

Chemistry
1 answer:
LUCKY_DIMON [66]2 years ago
3 0

Molar mass of acetone (CH_{3}) _{2} CO is 58.07914 g/mol

Elements presents in acetone is

Carbon, Hydrogen and Oxygen

Atomic mass of carbon is = 12.01 u

Atomic mass of Hydrogen = 1..00 u

Atomic mass of Oxygen = 15.99 u

So molar mass of acetone will be

= (CH_{3}) _{2} CO

= ( 12.01 + 3*(1.00))*2 + 12.01 + 15.99

=58.07914 g/mol

To learn more about Molar mass

brainly.com/question/22997914

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Hi, May I have Chem help please? Please Keep the answer as simple as possible because it is a for a bcr.
finlep [7]

Proton:

Mass = 1

Charge = +1

Location = Nucleus

Neutron:

Mass = 1

Charge = 0

Location = Nucleus

Electron:

Mass = 1/2000

Charge = -1

Location = Energy levels

6 0
3 years ago
an atmosphere is considered hazardous if it contains a hazardous gas in excess of 10 percent of the hazardous material's:
svlad2 [7]

Lower flammable limit means the lowest concentration of a material that will propagate a flame.

What is hazardous atmosphere?

It is an atmosphere that may expose employees to risk of death, incapacitation, impairment of ability to self-rescue, injury, or acute illness from one or more of following causes

  • Flammable gas, vapor, or mist in excess of 10 percent of lower flammable limit (LFL)
  • Airborne combustible dust at concentration that meets or exceeds its LFL

What is lower flammable limit?

  • It means the lowest concentration of a material that will propagate a flame.
  • The LFL is usually expressed as percent by volume of material in air (or other oxidant)
  • Atmospheres with concentration of flammable vapors at or above 10 percent of lower explosive limit (LEL) are considered hazardous when located in confined spaces.
  • However, atmospheres with flammable vapors below 10 percent of LEL are not necessarily safe. Such atmospheres are too lean to burn

Learn more about lower flammable limit at brainly.com/question/2456135

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6 0
1 year ago
describe in general terms an experiment to determine the molal freezing point depression constant kf of water. Assume the availa
Dvinal [7]
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point: 
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b -  molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.

6 0
3 years ago
Decreasing the temperature of the reaction 3H2 + N2<----->2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL
AURORKA [14]

3H_{2}+N_{2}⇔2NH_{3}

Decreasing the temperature of the reaction,the reaction shifts forward.

The explanation is given below.

Explanation:

If the temperature of the reaction mixture is increased,then the equilibrium will shift to decrease the temperature.

If the temperature of the reaction mixture is decreased,then the equilibrium will shift to increase the temperature.

During the formation of the ammonia,it gives off heat.So it is an exothermic reaction.

3H_{2}+N_{2}⇔2NH_{3}

A decrease in the temperature favors the reaction that is exothermic (the forward reaction)because it produces energy.Therefore,if the temperature is decreased,the yield of the ammonia increases.

<em>Therefore if the temperature is increased,the reaction shifts forward and the yield of the ammonia increases and it is an exothermic reaction.</em>

7 0
3 years ago
An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
8 0
3 years ago
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