Answer: The correct option is E.
Explanation: The reaction between aspirin (also known as acetylsalicylic acid) and sodium hydroxide is known as acid-base titration reaction.
By applying Unitary method, we get:
15.50mL of NaOH dissolves = 0.325 g of aspirin
So, 16.25 mL of NaOH will dissolve = = 0.341 g
Hence, the correct option is E.
Answer:
Number of moles = 0.153 mol
Explanation:
Given data:
Mass of sulfur = 5 g
Number of moles of sulfur atom = ?
Solution:
Formula:
Number of moles = mass/molar mass
Molar mass of sulfur is 32. 065g/mol.
By putting values,
Number of moles = 5 g/ 32.06 g/mol
Number of moles = 0.153 mol
Answer:
The correct answer is 16.61 grams methanol and 57.38 grams water.
Explanation:
The mole fraction (X) of methanol can be determined by using the formula,
X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)
X₁ = n₁/n₁ + n₂ = 0.14
n₁ / n₁ + n₂ = 0.14 ---------(i)
n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)
n₁ mole CH₃OH = 32.042 n₁ g
n₂ mole H2O = n₂ mole × 18.015 g/mol
n₂ mole H2O = 18.015 n₂ g
Thus, total mole number is,
32.042 n₁ + 18.015 n₂ = 74 ------------(ii)
From equation (i)
n₁/n₁ + n₂ = 0.14
n₁ = 0.14 n₁ + 0.14 n₂
n₁ - 0.14 n₁ = 0.14 n₂
n₁ = 0.14 n₂ / 1-0.14
n₁ = 0.14 n₂/0.86 ----------(iii)
From eq (ii) and (iii) we get,
32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74
n₂ (32.042 × 0.14/0.86 + 18.015) = 74
n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)
n₂ = 3.1854 mol
From equation (iii),
n₁ = 0.14/0.86 n₂
n₁ = 0.14/0.86 × 3.1854
n₁ = 0.5185 mol
Now, presence of water in the mixture is,
= 3.1854 mole × 18.015 gram per mole
= 57.38 grams
Methanol present in the mixture is,
= 0.5185 mol × 32.042 gram per mole
= 16.61 grams
In a 74.0 g aqueous solution of methanol with a mole fraction of 0.140, the mass of methanol is approximately 10.36 g and the mass of water is approximately 63.64 g.
The problem involves the calculation of the mass of the components of an aqueous solution of methanol (CH3OH). First, we need to know that the mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the mixture.
Given that the mole fraction of methanol is 0.140, this means that the rest of the solution (i.e., water) is 1 - 0.140 = 0.860. To find the mass of each component, we need to consider the total mass of 74.0 g.
The mass of methanol can be calculated as 74.0 g * 0.140 = 10.36 g. And the mass of water would be 74.0 g * 0.860 = 63.64 g.
So, in this aqueous solution, you have approximately 10.36 g of methanol and 63.64 g of water.
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Answer:
6
Explanation:
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A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. the final molarity of the dilute solution is 0.102 M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M₁) = 5.103 M
Volume of diluted solution (V₂) = 250 mL
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
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Answer:
0.102 M.
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 5 mL
Molarity of stock solution (M1) = 5.103 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (V2) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M1V1 = M2V2
5.103 × 5 = M2 × 250
25.515 = M2 × 250
Divide both side by 250
M2 = 25.515 / 250
M2 = 0.102 M
Thus, the molarity of the diluted solution is 0.102 M.
Answer:
19.6 J
Explanation:
The following data were obtained from the question:
Mass (m) of object = 2 Kg
Height (h) = 1 m
Workdone =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
Thus, we can obtain the workdone in lifting the object by using the following formula:
Workdone = mgh
Workdone = 2 × 9.8 × 1
Workdone = 19.6 J
Therefore, the workdone in lifting the object to height of 1 m is 19.6 J