Answer:
iodine
Explanation:
In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test. For example, if iodine is added to a peeled potato then it will turn black. Benedict's reagent can be used to test for glucose.
When water is boiled, the heat energy is transferred to the molecules of water, which begin to move more quickly. Eventually, the molecules have too much energy to stay connected as a liquid. When this occurs, they form gaseous molecules of water vapor, which float to the surface as bubbles and travel into the air.
The element iodine (I) is important for the fast and hastened metamorphosis of frog-tadpoles. As amphibians, the tadpoles can live in water and land but when they are born they are iodine-deficient. Tadpoles that do not receive ample amount of iodine become tadpoles until the end of their days.
Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
The balanced reaction equation for the reaction between CH₃OH and O₂ is
2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles 12 24
Reacted moles 12 18
Final moles - 6 12 24
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
= 12 mol x 3 / 2
= 18 mol
All of CH₃OH moles react with O₂.
Hence, the limiting agent is CH₃OH.
Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
