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Answer:
may be exchanged for equity securities.
Explanation:
Convertible bonds -
It refers to the type of bond which can easily be converted to stocks .
It refers to as the fixed - income debt security which can give the interest payments and can be converted to some predetermined number of equity shares and common stock , is referred to as convertible bonds .
The process of conversion can be done at any time period of the bond .
Hence , from the given information of the question ,
The correct answer is may be exchanged for equity securities.
Im sure that the answer is C, correct me if im wrong.
Answer:
[Acetic acid] = 0.07 M
[Acetate] = 0.13 M
Explanation:
pH of buffer = 5
pKa of acetic acid = 4.76
![pH=p_{Ka} + log\frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3Dp_%7BKa%7D%20%2B%20log%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now using Henderson-Hasselbalch equation
![5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}](https://tex.z-dn.net/?f=5%3D4.76%20%2B%20log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D)
![log\frac{[Acetate]}{[Acetic\;acid]} = 0.24](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%200.24)
![\frac{[Acetate]}{[Acetic\;acid]} = 1.74](https://tex.z-dn.net/?f=%5Cfrac%7B%5BAcetate%5D%7D%7B%5BAcetic%5C%3Bacid%5D%7D%20%3D%201.74)
....... (1)
It is given that,
[Acetate] + [Acetic acid] = 0.2 M ....... (2)
Now solving both the above equations
[Acetate] = 1.74[Acetic acid]
Substitute the concentration of acetate ion in equation (2)
1.74[Acetic acid] + [Acetic acid] = 0.2 M
[Acetic acid] = 0.2/2.74 = 0.07 M
[Acetate] = 0.2 - 0.07 = 0.13 M
Answer:
4.87g
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Mass of solution = 0.35kg
Molality = 0.238 m
Mass of NaCl =..?
Step 2:
Determination of the number of mole of NaCl in the solution.
Molality of a solution is simply defined as the mole of solute per unit kg of the solvent. It is given as:
Molality = mol of solute /mass of solvent (kg)
With the above formula, we calculate the mole of NaCl present in the solution as follow:
Molality = mol of solute /mass of solvent (kg)
0.238 = mol of NaCl /0.35
Cross multiply
mol of NaCl = 0.238 x 0.35
mol of NaCl = 0.0833 mol
Step 3:
Determination of the mass of NaCl in 0.0833 mol of NaCl.
This is illustrated below:
Number of mole NaCl = 0.0833 mol
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass of NaCl =..?
Mass = number of mole x molar Mass
Mass of NaCl = 0.0833 x 58.5
Mass of NaCl = 4.87g
Therefore, 4.87g of NaCl is contained in the solution.
