Yes because some work harder than others to get their credit for developing the periodic table
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
You can answer this question by only searching the element in the periodic table.
The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).
The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
Answer : The initial temperature of system 2 is, 
Explanation :
In this problem we assumed that the total energy of the combined systems remains constant.
The mass remains same.
where,
= heat capacity of system 1 = 19.9 J/mole.K
= heat capacity of system 2 = 28.2 J/mole.K
= final temperature of system =
= initial temperature of system 1 =
= initial temperature of system 2 = ?
Now put all the given values in the above formula, we get
Therefore, the initial temperature of system 2 is, 