Answer:
We show the data that is collected by in this way :-
Determine What Information You Want to Collect. The first thing you need to do is choose what details you want to collect. ...
Set a Timeframe for Data Collection. ...
•Determine Your Data Collection Method. ...
Collect the Data.
Analyze the Data and Implement Your Findings
The solution to the problem is as follows:
This is mostly stoichiometry with a little bit of substitution into the acid dissociation equation.
You need intial moles of acid and salt = 0.025 x 0.16 = 0.004 mol
You need moles HA after 8ml of 0.21 M KOH added = 0.004 - (0.008 x 0.21) = 0.00232 mol HA
moles of salt = moles of acid reacted = (0.008 x 0.21) = 0.00168 mol A-
Now work out concentrations based on total volume = 0.025 + 0.008 = 0.033 dm3
[HA] = 0.00232/0.033 = 0.0703 M
[A-] = 0.00168/0.033 = 0.0509 M
ka = [H+][A-]/[HA]
1.74 x 10-5 = [H+] * 0.0509/0.0703
[H+] = 2.4 x 10-5
pH = 4.62
<span>For the 40 ml addition there is no acid left over, so it's just calculated from excess base.
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Answer:
1. unless you live right around the block, distance from home to school should be in kilometers.
2. centimeters
3. millimeters
4. meters (the average is about 2 meters)
Answer: 4 molL-1
Explanation:
Detailed solution is shown in the image attached. The number of moles of NaCl is first obtained. Since the molarity must be in units of molL-1, the volume is divided by 1000 and the formula stated in the solution is applied and the answer is given to one significant figure.
Answer:
carbonyl (-C=O)
hydroxyl (-OH)
Explanation:
Sugars have a carbonyl group that interacts with an hydroxyl group that forms a ring structure when the dry molecule is placed in water. Let take a look at sucrose for example.
Sucrose have the molecular formula of C₁₂H₁₂O₁₁. Its preparation is gotten from a juice of sugar cane. In the process of extracting the juice from sugar cane, the extraction process is done with water usually 80°C. The solution is then treated (purification) afterwards with slaked lime and carbon(iv)oxide. During this stage, the (-C=O) group interfere with the OH group in order to form a ring structure.