A shiny chunk of metal is found to have a mass of 37.28g. The metal is dropped into a graduated cylinder which contains 20.0 mL
of water. The volume of the cylinder rises to 34.0 mL. Determine the density of the metal. What might the metal be made of?
1 answer:
Answer: The density of the material is 2.66 g/mL and it is likely this is made of Aluminum
Explanation:
The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = . Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:
This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.
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<u>Answer:</u> The pH of the resulting solution will be 3.60
<u>Explanation:</u>
Molarity is calculated by using the equation:
......(1)
We are given:
Molarity of formic acid = 0.100 M
Molarity of potassium formate = 0.100 M
Volume of solution = 420 mL = 0.420 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
Molarity of KOH = 1.00 M
Volume of solution = 7 mL = 0.007 L
Putting values in equation 1, we get:
The chemical equation for the reaction of formic acid and KOH follows:
I: 0.042 0.007 0.042
C: -0.007 -0.007 +0.007
E: 0.035 - 0.049
Volume of solution = [420 + 7] = 427 mL = 0.427 L
To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:
.......(2)
Given values:
Putting values in equation 2, we get:
Hence, the pH of the resulting solution will be 3.60
Answer : The value of equilibrium constant (K) is, 0.004
Explanation :
First we have to calculate the concentration of
and,
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 1.2 0 0
At eqm. (1.2-2x) 2x x
As we are given:
Concentration of at equilibrium = x = 0.1 M
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
Thus, the value of equilibrium constant (K) is, 0.004