Answer:

Explanation:
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 28.01 17.03
N₂ + 3H₂ ⟶ 2NH₃
m/g: 240.0
(a) Moles of NH₃

(b) Moles of N₂

(c) Mass of N₂

In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
Answer:
a) 32.09 kPa
b) 32.09 kPa
Explanation:
Given data:
rate constant 
initial pressure is = 32.1 kPa
half life of A is calculated as



for calculating pressure we have follwing expression


a) 
b) 
Answer:
8.61 mL of the HCl solution
Explanation:
The reaction that takes place is:
- 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
From the given mass of Mg(OH)₂, we can calculate <u>the moles of HCl that are neutralized</u>:
- 4x10² mg = 400 mg = 0.400g
- 0.400g Mg(OH)₂ ÷ 58.32g/1mol = 6.859*10⁻³ mol Mg(OH)₂
- 6.859*10⁻³ mol Mg(OH)₂ *
3.429x10⁻³ mol HCl
Finally, to calculate the volume of an HCl solution, we need both the moles and the concentration. We can <u>calculate the concentration using the pH value</u>:
= [H⁺]
- 0.0398 M = [H⁺] = [HCl] *Because HCl is a strong acid*
Thus, the volume is:
- 0.0398 M = 3.429x10⁻³mol HCl / Volume
- Volume = 8.616x10⁻³ L = 8.62 mL
Answer:
Lewis structure for isomers of butane has been given below.
Explanation:
Butane is a saturated alkane with molecular formula
. Due to different positing of methyl groups, positional isomerism exists in butane.
Butane has two positional isomers with same molecular formula. One is n-butane and another one is isobutane. Lewis structures of these two isomers have been given below.