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3 years ago

What is the mass of cyclohexane solvent, in kg, if 9.76 mL are used and the cyclohexane has a density of 0.779 g/mL?

Chemistry

1 answer:

nordsb [41]3 years ago

3 0

Answer:

0.0076kg

Explanation:

To get the mass, we use the relation among density, mass and volume.

Mass = density * volume

Here mass? , density = 0.779g/ml , volume = 9.76ml

Mass = 9.76 * 0.779 = 7.60g

Answer is wanted in kg so we divide by 1000. This is 7.60/1000 = 0.0076kg

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3 years ago

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If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?

Alex73 [517]

Answer:

0.189 g.

Explanation:

This problem is an application on Henry's law.

Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.

Solubility of the gas ∝ partial pressure

If we have different solubility at different pressures, we can express Henry's law as:

S₁/P₁ = S₂/P₂,

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

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8 0

3 years ago

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

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3 years ago

What is the mole fraction of ethanol in a solution of 3.00 moles of ethanol and 5.00 moles of water?

Rus_ich [418]

If theres a mixture of components we can calculate the mole fraction
mole fraction can be calculated as follows
mole fraction of component = $\frac{number of moles of component}{total number of moles of all components }$
number of moles of ethanol - 3.00 mol
total number of moles in mixture - 3.00 + 5.00 = 8.00 mol

mole fraction of ethanol = $\frac{3.00 mol}{8.00 mol }$
mole fraction of ethanol is 0.375

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3 years ago

Draw the structure of the major organic product isolated from the reaction of 1-hexyne with sodium amide in liquid ammonia follo

77julia77 [94]

Answer:check explanation and attached file/picture

Explanation:

Alkynes are hydrocarbons and they are very acidic because of hybridization effect(the more acidic the s-character is, the more acidic it is going to be).

The reaction of 1-hexyne with sodium amide in liquid ammonia is a form of deprotonation 'reaction' to form acetylide. Due to the acidic nature of the terminal hydrogen atom, terminal alkynes do form metallic derivatives by the replacement of the terminal hydrogens. The equation of Reaction is given below.

C6H10 + NaNH2( in liquidNH3) ------------> C6H9Na + H-NH2.

The acetylide is a bae and a very good nucleophile.

The reaction is then followed by the addition of 1-bromobutane. This reaction is used for the production of longer chain alkynes. The equation of Reaction is attached in the picture.

3 0

3 years ago

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