Answer:
Mass = 14.0 g
Explanation:
Given data:
Mass of mercury nitrate = 22.17 g
Mass of mercury formed = ?
Solution:
Chemical equation:
Hg(NO₃)₂ + 2K → 2KNO₃ + Hg
Number of moles of mercury nitrate:
Number of moles = mass/molar mass
Number of moles = 22.17 g / 324.6 g/mol
Number of moles = 0.07 mol
Now we will compare the moles of Hg(NO₃)₂ and mercury.
Hg(NO₃)₂ : Hg
1 : 1
0.07 : 0.07
Mass of mercury:
Mass = number of moles × molar mass
Mass = 0.07 mol × 200.6 g/mol
Mass = 14.0 g
<h3><u>Answer; </u></h3>
=10.38 moles KOH
<h3><u>Explanation</u>;</h3>
The balanced equation.
6KOH + Al2(SO4)3 --> 3K2SO4 + 2Al(OH)3
From the equation;
1 mole of aluminum sulfate requires 6 moles of potassium hydroxide.
Moles of Aluminium sulfate; 1.73 moles
Moles of KOH;
1 mol Al2(SO4)3 : 6 mol KOH = 1.73 mol Al2(SO4)3 : x mol KOH
Thus; x = (6 × 1.73)
<u> =10.38 moles KOH </u>
Chechnya. 345. Gcbjshjkfs
Answer:
r = 3.61x
M/s
Explanation:
The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.
r = k.![[S2O2^{-8} ]^{x} x [I^{-} ]^{y}](https://tex.z-dn.net/?f=%5BS2O2%5E%7B-8%7D%20%5D%5E%7Bx%7D%20x%20%5BI%5E%7B-%7D%20%5D%5E%7By%7D)
K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :
x = 1
Now, to find the coefficient y let's do the same for the experiments 1 and 3:
y = 1
Now, we need to calculate the constant k in whatever experiment. Using the first :
k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]
Using the data given,
r = 
r = 3.61x M/s
