Determine how many mm of Hg are equal to 5.3 atm of pressure

tensa zangetsu [6.8K]

Answer:

We will produce 8.2 grams of acetylsalicylic acid

Explanation:

step 1: Data given

Mass of salicylic acid, = 6.3 grams

Molar mass salicylic acid = 138.12 g/mol

Molar mass of acetylsalicylic acid = 180.158 g/mol

Step 2: The balanced equation

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

Step 3: Calculate moles salicylic acid

Moles salicylic acid = mass salicylic acid / molar mass salicylic acid

Moles salicylic acid = 6.3 grams / 138.12 g/mol

Moles salicylic acid = 0.0456 moles

Step 4: Calculate moles acetylsalicylic acid

Since the mole ratio is 1 to 1

For 0.0456 moles salicylic acid we'll have 0.0456 moles acetylsalicylic acid

Step 5: Calculate mass acetylsalicylic acid

MAss acetylsalicylic acid = moles acetylsalicylic acid * molar mass acetylsalicylic acid

MAss acetylsalicylic acid = 0.0456 moles * 180.158 g/mol

Mass acetylsalicylic acid = 8.2 grams

We will produce 8.2 grams of acetylsalicylic acid

4 0

3 years ago

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .