Answer:
Mass % of hydrogen = 20.9 %
Explanation:
Given data:
Mass of hydrogen = 0.485 g
Mass of water = 2.32 g
Mass percent of hydrogen = ?
Solution:
Mass % of hydrogen = mass of hydrogen / mass of water × 100
Mass % of hydrogen = 0.485 g/ 2.32 g × 100
Mass % of hydrogen = 0.209 × 100
Mass % of hydrogen = 20.9 %
<span>1.)One way to limit iron corrosion is to coat the iron with a metal that:
A.)will never undergo corrosion
</span>
The balanced chemical reaction is:
N2 + 3H2 = 2NH3
We are given the amount of ammonia formed
from the reaction. This is where we start our calculations.
0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol
H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2
Answer:
It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8
Explanation:
The equation that represents a first-order kinetics is:
Ln ([A] / [A]₀] = -kt
<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>
<em />
As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8
Replacing:
Ln ([A] / [A]₀] = -kt
Ln (1/8) = -1.57x10⁷s⁻¹*t
t = 1.32x10⁻⁷s
<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>
MgCl₂ & PbSO₄ is the correct product for the given double replacement reaction.
<h3>What is double displacement reaction?</h3>
In the double displacement reaction displacement of two substrate will take place among two reactants and formation of products takes place.
In the question, PbCl₂ reacts with MgSO₄ and displacement between the cations takes place and formation of MgCl₂ and insoluble precipitate PbSO₄ is formed.
Given reaction is represented as:
PbCl₂ + MgSO₄ → MgCl₂ + PbSO₄
According to the activity series, reactivity of Magnesium is more as compared to the lead atom. So lead is displaced by the magnesium atom and form the above given products.
Hence, option (3) is correct i.e. MgCl₂ & PbSO₄.
To know more about double displacement reaction, visit the below link:
brainly.com/question/26413416