Answer:
Volume = 35.2×220×6.0 = 46464 centimeters³
Explanation:
In 1869 he published a table of the elements organized by increasing atomic mass.
Mendeleev is called the "father of the modern periodic table
stated that if the atomic weight of an element caused it to be placed in the wrong group, then the weight must be wrong. (He corrected the atomic masses of Be, In, and U)
was so confident in his table that he used it to predict the physical properties of three elements that were yet unknown.
After the discovery of these unknown elements between 1874 and 1885, and the fact that Mendeleev's predictions for Sc, Ga, and Ge were amazingly close to the actual values, his table was generally accepted.
However, in spite of Mendeleev's great achievement, problems arose when new elements were discovered and more accurate atomic weights determined.
STOP DELETING MY COMMENTS PEOPLE PLEASE!!!!!!!!!
Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
Easy peasy! All we need to do is plug this formula into our calculator:
-log(M)
So, we'd plug in -log(.2), which is 0.7 :)
Answer: <span>option C. Measure the four products separately, but using the
same scale each time.
</span>
Justification:
Review the choices given one by one.
<span> A. Measure the product separately
on four different scales.
</span><span>
</span><span>
</span><span>He should not include other variables in the experiment. Using four different scales migh give different results due to differences on the scales.
</span><span>
</span><span>B. Measure all of the product together and
divide by four.
</span><span>
</span><span>
</span><span>By doing this he will just obtain an average of the amount produced, but in this way he cannot determine the production individually (with each catalyst).
</span><span>
</span><span>
</span><span>C. Measure the four products separately, but using the
same scale each time.
</span><span>
</span><span>
</span><span>Indeed doing this he will be able to compare and rank the efficiency of each catalyst.
</span><span>
</span><span>
</span><span>D. Measure the amount of sodium chloride in one
sample every 10 seconds.
</span>
He cannot do this without affecting at the same time the evolution of the reaction, and of course by doing it on one sample only he will not be able to compare.
