The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
Answer:
13.7 moles of O₂ are needed
Explanation:
In order to find the moles of reactants that may react to make the products we need to determine the reaction:
Reactants are hydrogen and oxygen
Product: Water
2 moles of hydrogen can react to 1 mol of oxygen and produce 2 moles of water.
Balanced reaction: 2H₂(g) + O₂(g) → 2H₂O(l)
If 2 moles of hydrogen need 1 mol of oxygen to react
Therefore, 27.4 moles of H₂ must need (27.4 .1) / 2 = 13.7 moles of O₂
We can calculate how long the decay by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span><span>From the half-life data, we can calculate for k.
</span>
1/2(Ao) = Ao e^-k(30)
<span>k = 0.023
</span>
0.04Ao = Ao e^0.023(t)
<span>t = 140 sec</span>
We all know, Density = Mass/ volume.
When volume increases and the mass remains the same , the density will decrease considerably.
