Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
The answer is Solid.
This is on account of the substances that develop a solid are packed in a settled, firmly pressed geometric plan.
For the neutralization process: an acid acts as a donor and donates protons to the base. On the other hand, the base acts as an acceptor and accepts the transferred protons. In a nutshell, neutralization is mainly proton transfer process.
As for the redox process: the oxidized material usually transfers electrons to the reduced material. In a nutshell, redox is mainly electron transfer process.
They are called isotopes.
Isotopes have the same number of electrons and protons in their unionized state. They differ in the number of neutrons. The first and simplest example is hydrogen.
The most common hydrogen has
1 proton
1 electron and
0 neutrons
It has 2 cousins
1 proton
1 electron
1 neutron
And
1 proton
1 electron
2 neutrons.
Most elements have some differences in the number of neutrons present in their nuclei. Cesium and Xenon have the most number of isotopes. Each has 36. You wonder how the atoms are held together.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
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