The starting substances in a chemical reaction are called reactants - they are written on the left side of a chemical equation.
Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7
Given data :
mass of aspirin = 640 mg = 0.640 g
volume of water = 10 ounces = 0.295735 L
molar mass of aspirin = 180.16 g/mol
moles of aspirin = mass / molar mass = 0.00355 mol
<h3>Determine the pH of the solution </h3>
First step : <u>calculate the concentration of aspirin</u>
= moles of Aspirin / volume of water
= 0.00355 / 0.295735
= 0.012 M
Given that pKa of Aspirin = 3.5
pKa = -logKa
therefore ; Ka = 
= 
From the Ice table
= ![\frac{x + H^+}{[aspirin]}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%20%2B%20H%5E%2B%7D%7B%5Baspirin%5D%7D)
= 
given that the value of Ka is small we will ignore -x
x² = 
x = 
Therefore
[ H⁺ ] =
given that
pH = - Log [ H⁺ ]
= - ( -3 + log 1.948 )
= 2.71 ≈ 2.7
Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7
Learn more about Aspirin : brainly.com/question/2070753
Concentration of unknown acid is 0.061 M
Given:
Concentration of NaOH = 0.125 M
Volume of NaOH = 24.68 mL
Volume of acid solution = 50.00 mL
To Find:
Concentration of the unknown acid
Solution: Concentration is the abundance of a constituent divided by the total volume of a mixture. The concentration of the solution tells you how much solute has been dissolved in the solvent
Here we will use the formula for concentration:
M1V1 = M2V2
0.125 x 24.68 = 50 x M2
M2 = 0.125 x 24.68 / 50
M2 = 0.061 M
Hence, the concentration of unknown acid is 0.061 M
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