The empirical formula is C₇H₆O₂.
Assume that you have 100 g of the compound.
Then you have 68.84 g C and 4.962 g H.
Mass of O = (100 – 68.84 – 4.962) g = 26.20 g O.
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
<u>Element</u> <u>Mass/g</u> <u>Moles</u> <u>Ratio</u> <u> ×2</u> <u>Integers</u>
C 68.84 5.732 3.501 7.001 7
H 4.962 4.923 3.006 6.012 6
O 26.20 1.638 1 2 2
The empirical formula is C₇H₆O₂.
Answer:
This question is incomplete; the complete part is:
A) All cell have a cell wall.
B) All cell arise from pre-existing cells.
C) All cell are capable of photosynthesis.
D) All cell can develop into any other type of cell.
The answer is B
Explanation:
The commonly known universal theory proposed in 1838 took the contribution from three remarkable scientists namely: botanist Matthias Schleiden, anatomist Theodor Schwann and biologist Rudolph Virchow. According to the question, Mathias discovered that all plants are made of cells, Schwann determined that all animals are made of cells while Virchow determined that all living things are composed of cells.
However, in addition to Virchow's discovery, he also discovered and proposed that "All cell arise from pre-existing cells", which till date forms part of the three components of the cell theory. The three parts are:
- Cell is the fundamental and basic unit of all living things.
- All living things are made up of one or more cells
- All cells arise from pre-existing cells
Answer:
The age of the sample is 4224 years.
Explanation:
Let the age of the sample be t years old.
Initial mass percentage of carbon-14 in an artifact = 100%
Initial mass of carbon-14 in an artifact = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final mass percentage of carbon-14 in an artifact t years = 60%
Final mass of carbon-14 in an artifact = ![[A]=0.06[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D0.06%5BA_o%5D)
Half life of the carbon-14 = 
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%5Ctimes%20t%7D)
![0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}](https://tex.z-dn.net/?f=0.60%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7B5730%20year%7D%5Ctimes%20t%7D)
Solving for t:
t = 4223.71 years ≈ 4224 years
The age of the sample is 4224 years.
You can determine it by paying attention to the unique characteristics that could only be found at heart's tissue, such as :
- looks striated or stripped
- The bundles are breached like tree but connected at both ends
hope this helps
78.6 g (1mol/60.1 g)= 1.31 moles of isopropanol
