Answer:
Percentage composition of oxygen = 29.69%
Percentage composition of fluorine = 70.37
Explanation:
Given:
Mass of compound = 0.432 grams
Mass of oxygen = 0.128 grams
Find:
Percentage composition of oxygen
Percentage composition of fluorine
Computation:
Percentage composition of oxygen = [0.128/0.432]100
Percentage composition of oxygen = 29.69%
Percentage composition of fluorine = [(0.432 - 0.128)/0.432]100
Percentage composition of fluorine = 70.37
The structure of this oligopeptide is attached:
It consists of arginine - alanine - phenylalanine and leucine
From the picture we can see that N terminus charge is +1
Also Arginine is considered as basic amino acid with charge +1
C terminus charge is -1
At pH = 8, the whose pKa are less than pH will be deprotonated and the net charge will be +1
So the net charge of this oligopeptide is +1 at pH = 8
Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
Answer:
4.43L is final volume of the ballon
Explanation:
Avogadro's law of ideal gases states that <em>equal volumes of gases, at the same temperature and pressure, have the same number of molecules</em>.
The formula is:

Where V and n are volume and moles of the gas in initial and final conditions.
If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

V₂ = <em>4.43L is final volume of the ballon</em>