Answer:
5.4 tonnes.
Explanation:
The first step is to find the molar mass of Al2O3. Aluminum has a molar mass of about 27 and oxygen has a molar mass of about 16, so 2(27)+3(16)= 102g/mol=0.102kg/mol. 10200kg/0.102kg/mol=100,000 moles of Al2O3 in 10.2 tonnes. Multiplying this by the molar mass of the two aluminums, you get a total of 54*100,000=5400000g=5400kg=5.4 tonnes. Hope this helps!
Answer:
150mL
Explanation:
First, let us calculate the volume of the diluted solution. This is illustrated below:
Information obtained from the question include:
V1 (initial volume) = 50mL
C1 (initial concentration) = 1M
C2 (final concentration) = 0.25M
V2 (final volume) =.?
The volume of the diluted solution can be obtained using the dilution formula C1V1 = C2V2 as shown below:
C1V1 = C2V2
1 x 50 = 0.25 x V2
Divide both side by 0.25
V2 = 50/0.25
V2 = 200mL
The volume of the diluted solution is 200mL.
Now, to obtain the volume of the water added to the solution, we'll simply subtract the initial volume from the final volume as shown below
Volume of water added = V2 - V1
Volume of water added = 200 - 50
Volume of water added = 150mL
Therefore, the volume of water added is 150mL
Answer:
C would be the most best choice for this question.
Thank you.
If you have any questions, feel free to leave a comment and I will get back to it as soon as I can.
-The Spectre
Explanation:
20000 nanograms sorry i cant answer the second question
Solution :
The buffer is the one that contains weak acid (the 
is nearly equal to the required pH) and the salt of its conjugate base.
The values of the given weak acids are as follows :
HCOOH : = 3.744
HCN : = 9.21
Since the required pH is 5.03, the suitable buffer is the mixture of the acetic acid and the salt of its conjugate base.
Let suppose the volumes of 
and 
are x and y mL respectively.
So the total volume of the buffer is 1000 mL.
∴ x+ y = 1000 ................(1)
Writing the Henderson-Hasselbalch equation for the given buffer solution :
![$pH = pK_a + \log \ \frac{[CH_3COONa]}{[CH_3COOH]}$](https://tex.z-dn.net/?f=%24pH%20%3D%20pK_a%20%2B%20%5Clog%20%5C%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%24)
.............(2)
y = 1.9 x
Substituting the values of y in equation (1), we get
x + 1.9 x = 1000
x = 345
Putting the value of x in (1), we get
345 + y = 1000
y = 655
Therefore the volume of is 345 mL and the volume of is 655 mL.
