Answer:
104.969 amu.
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 107.977 amu
Abundance (A%) = 0.1620%
Isotope B:
Mass of B = 106.976 amu
Abundance (B%) = 1.568%
Isotope C:
Mass of C = 105.974 amu
Abundance (C%) = 47.14%
Isotope D:
Mass of D = 103.973 amu
Abundance (D%) = 51.13%
Average atomic mass =?
The average atomic mass of the element can be obtained as follow:
Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]
Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]
= 0.175 + 1.677 + 49.956 + 53.161
= 104.969 amu
Therefore, the average atomic mass of the element is 104.969 amu.
Answer:
5.42g, 71.77%
Explanation:
First, we have to write out the balanced chemical equation. The unbalanced equation can be written as “SO2+O2 -> SO3” and to balance it, we can see that having two mols of SO2 and two mols of SO3 will make each side have the same amount of mols per element on each side. So the balanced chemical equation is “2SO2 + O2 -> 2SO3”
Now, we want to solve for the theoretical yield in grams of SO3. To do this, we have to use dimensional analysis. We convert g SO2 into mols SO2 using the molar mass of the elements. Then we convert mols of SO2 into mols of SO3 using the balanced equation. Once we’ve done that, we can convert mols of SO3 into grams of SO3.
You should know how to look up the molar mass of elements on the periodic table by now. Find the masses and set up the terms so they cancel like so:
Doing the math, we get 5.42g so3 as the theoretical yield. This is the most amount that you could ever get if the world was a perfect place. But alas, it isn’t and mistakes are gonna happen, so the number is going to be less than that. So the best we can do, is to figure out the percent yield that we got.
In a lab scenario, this was calculated to be 3.89 g as stated by the problem. The percent composition formula is
and plugging the numbers into it, we get:
make sure to follow the decimal/significant figure rules of your instructor, but only round at the end. My professor didn't care too much thankfully, but some professors do
Answer:
Average density for method A = 2.4 g/cm³
Average density for method B = 2.605 g/cm³
Explanation:
In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):
Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6
Average = 9.6/4 = 2.4 g/cm³
Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420
Average = 10.420/4 = 2.605 g/cm³
Answer:
Hydrogen
Explanation:
Just to provide some background, an element is a pure substance consisting of only one type of atom. An atom is the smallest constituent of matter. All elements are comprised of a single type of atom (e.g., gold is composed of gold atoms, helium of helium atoms, phosphorus phosphorus, and so on).
A molecule is a group of two or more atoms. They can be the same atom (homonuclear), such as or different atoms (heteronuclear).
Some examples of homonuclear molecules include:
Hydrogen (H2)
Nitrogen (N2)
Phosphorus (P4)
Some examples of heteronuclear molecules include:
Carbon dioxide (CO2)
Sulfuric acid (H2SO4)
Methane (CH4)
