A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remain
1 answer:
A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
Explanation:
Ore contains ---- 65.2%
Mass% of titanium in TiO2 can be calculated as shown below:
Given 10.0 metric tons of titanium is required.
The mass of ore that should be processed can be calculated as shown below:
Mass of Ti = ore x TiO2 % x Ti mass %
10.0 x 1000 kg = M (mass of ore) x (65.2/100) x (59.9/100) Ti
=>M=(10.0 metric tons) / (0.652 x 0.599)
=>M=25.6 metric tons
Hence, the mass of ore required is 25.6 metric tons.
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Answer: 0.20 M
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1.40 M
= volume of stock solution = 72.0 ml
= molarity of diluted solution = m
= volume of diluted solution = 248 ml
Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.
According to the dilution law,
where,
= molarity of stock solution = 0.41 M
= volume of stock solution = 124 ml
= molarity of diluted solution = m
= volume of diluted solution = (124 +133) ml = 257 ml
Thus the final concentration of the solution is 0.20 M.