Answer:
24.7 amu
Explanation:
An isotope is when an element can have different number of neutrons but they have same number of protons.
In order to calculate the average atomic mass with the given information do the following operations:
First change de percentages to fractional numbers, divide by 100.
I like to make a table, to organize all data and I believe is easier to understand.
65/100 = 0.65
35/100 = 0.35
% fraction
65.0 0.65
35.0 0.35
total100.0 1
Now multiply each mass with their corresponding fraction
24 (0.65) = 15.6
26 (0.35) = 9.1
% fraction uma uma
65.0 0.65 24 15.6
35.0 0.35 26 9.1
total100.0 1 24.7
Finally you add the resulting mass and the units will be in uma.
15.6+9.1 = 24.7
Therefore the average atomic mass of this element will be 24.7 uma.
Check the table in the document attached
Producers are the foundation of every food web in every ecosystem—they occupy what is called the first tropic level of the food web. The second trophic level consists of primary consumers—the herbivores, or animals that eat plants. At the top level are secondary consumers—the carnivores and omnivores who eat the primary consumers. Ultimately, decomposers break down dead organisms, returning vital nutrients to the soil, and restarting the cycle. Another name for producers is autotrophs, which means “self-nourishers.” There are two kinds of autotrophs. The most common are photoautotrophs—producers that carry out photosynthesis. Trees, grasses, and shrubs are the most important terrestrial photoautotrophs. In most aquatic ecosystems, including lakes and oceans, algae are the most important photoautotrophs.
Explanation:
In this reaction, the reactants are Li and N2. The product is Li3N
So we have;
Li + N2 → Li3N
Upon balancing, we have;
6Li + N2 → 2 Li3N
The sum of the coefficients is 6 + 1 + 2 = 9
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
