Answer:
8.8 yards per second
Explanation:
50 - 20 = 30 yards
30 yards/3.4 seconds = 8.8235
Answer:
F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)
Explanation:
Our reactants are:
F₂ → Fluorine gas, a dyatomic molecule
FeI₂ → Iron (II) iodine
Our products are:
I₂ → Iodine
FeF₂ → Iron (II) fluoride
Then, the reaction is:
F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)
We see it is completely balanced.
Answer:
Mass = 15.20 g of KCl
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
2 KClO₃ = 2 KCl + 3 O₂
Step 1: Calculate moles of KClO₃ as;
Moles = Mass / M/Mass
Moles = 25.0 g / 122.55 g/mol
Moles = 0.204 moles
Step 2: Find moles of KCl as;
According to equation,
2 moles of KClO₃ produces = 2 moles of KCl
So,
0.204 moles of KClO₃ will produce = X moles of KCl
Solving for X,
X = 2 mol × 0.204 mol / 2 mol
X = 0.204 mol of KCl
Step 3: Calculate mass of KCl as,
Mass = Moles × M.Mass
Mass = 0.204 mol × 74.55 g/mol
Mass = 15.20 g of KCl
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
Answer:
Solid: calcium and potassium
Liquid: mercury and bromine
Gaseous: oxygen and fluorine
Explanation:
Matter exists in three different states namely: solid, gaseous and liquid. Elements that are found in nature are classified as matter. Since, these elements are grouped as matter, they can either be found as either solids, liquids or gases at normal temperature and pressure.
At normal temperature (20°C or 293K) and pressure (1 atm), the following elements are found to exists in the respective state of matter:
Solid: Calcium (Ca) and potassium (K)
Liquid: Mercury (Hg) and bromine (Br)
Gaseous: Oxygen (O) and fluorine (F)
