Use a strip of paper covered in PH indicating dye.
Answer:
H2O> NH3> CH4
Explanation:
According to valence shell electron pair repulsion theory (VSEPR), bond angles and repulsion of electron pairs depends on the nature of electron pairs on the central atom of the molecule. Lone pairs cause more repulsion (and distortion of bond angles) than bond pairs). Lone pair- lone pair repulsion is greater than lone pair bond pair repulsion.
Water contains two lone pairs on oxygen hence it experiences the greatest repulsion. Ammonia has only one lone pair on nitrogen hence there is lesser repulsion between lone pairs and bond pairs. Methane possess only bond pairs of electrons hence it has the least repulsion.
<h3>
Answer:</h3>
Chlorine gas (Cl₂)
<h3>
Explanation:</h3>
- According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
- Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
- Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.
In this case we are given gases, CI₂
, H₂,He and Ne.
- We are required to identify the gas that will diffuse at the slowest rate.
- In other words we are required to determine the heaviest gas.
Looking at the molar mass of the gases given;
Cl₂- 70.91 g/mol
H₂- 2.02 g/mol
He - 4.00 g/mol
Ne- 20.18 g/mol
Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.
Moles of Bromine produced = 9 moles
<h3>Further explanation</h3>
Given
9 moles of Chlorine gas
Word equation
Required
Moles of Chlorine produced
Solution
We change the word equation into a chemical equation (with a formula)
Aluminum bromide reacts with chlorine gas to produce Aluminum chloride and bromide gas
2AlBr₃+3Cl₂⇒2AlCl₃+3Br₂
moles Cl₂ = 9
Maybe you mean, <em>how many moles of Bromine can we produce?</em>
From equation, mol ratio Cl₂ : Br₂ = 3 : 3, so mol Br₂=mol Cl₂=9 moles
Reaction.-
2H2 + O2 ---> 2H2O
Given,
Mass of Oxygen - 25.0 g
Mass of Hydrogen - 42.0 g
now we have to calculate moles
moles of O2 = 25/16 = 1.5625 moles
Moles of H2 = 42/2 = 21 moles
Hydrogen gas is in excess,
now using reaction
1 mole of oxygen gas forms 2 moles of H2O
1.5625 mole of Oxygen gas will form x mole of H20
x = 3.125 moles
molar mass of H2O is 18g
mass of 3.125moles of water = 3.125 × 18 = 56.25 g.
Hence the mass of water produced when 25.0g of oxygen gas react with 42.0g of hydrogen gas is 56.25 grams
