Answer:
We would need 10 mL of the concentrated CaCl₂ stock solution, and 30 mL of water.
Explanation:
To solve the question asked we can use the C₁V₁=C₂V₂ equation, where:
We <u>solve for V₁</u>:
We would need 10 mL of the concentrated CaCl₂ stock solution, and (40-10) 30 mL of water.
 
        
             
        
        
        
The energy needs of a cell is very important because they need energy to breakdown,synthesis, and transport nutrients and molecules so an living organism can survive just like how the city needs energy to power its cities so it’s people can do what they need to do to survive. HOPE IT HELPS
        
             
        
        
        
Answer: Hello,Mass ratio charge is.....................
Explanation:  In mass spectroscopy, the mass-to-charge ratio (symbols: m/z, m/e) of a cation is equal to the mass of the cation divided by its charge. ... The mass of the molecular ion is equal to the molecular weight of the compound. Thus, the mass-to-charge ratio of the molecular ion is equal to the molecular weight of the compound.
 
        
             
        
        
        
Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻)
= 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) =
5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.
        
             
        
        
        
For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.
n(Cd) / n(Fe)=1
<h3>What is the density of the nucleus 112 48cd?</h3>
Generally, the equation for the density  is mathematically given as
d=\frac{A}{4/3}\piR^3
Therefore
n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3
n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3
n(Cd) / n(Fe)=1
In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same
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