Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
It can tell us how old they are and how and where they moved in the past.
Answer:
F = 800 N
Explanation:
Given data:
Mass = 80 Kg
Acceleration = 10 m/s²
Force = ?
Solution:
Formula:
<em>F = m × a
</em>
F = force
m = mass
a = acceleration
Now we will put the values in formula:
<em>F = m × a
</em>
F = 80 kg <em>× </em>10 m/s²
F = 800 kg.m/s²
kg.m/s² = N
F = 800 N
<h3><u>Answer;</u></h3>
= 3032.15 kPa
<h3><u>Explanation;</u></h3>
Using the equation;
PV = nRT , where P is the pressure,. V is the volume, n is the number of moles and T is the temperature and R is the gas constant, 0.08206 L. atm. mol−1.
Volume = 7.5 L, T = 274 +273 = 547 K, N = 5 moles
Therefore;
Pressure = nRT/V
= (5 × 0.08206 × 547)/7.5 L
= 29.925 atm
But; 1 atm = 101325 pascals
Hence; Pressure = 3032150.63 pascals
<u>= 3032.15 kPa</u>
