Answer:
The name of that compound is Nickel(||)iodide
Answer:
E) Silver, mass = 420 g, d = 10.5 g/cm.
Explanation:
Hello,
In this case, since the density is defined by:

Given the mass and density, we compute the volume via:

Thus, for each vase we have:
A) 
B) 
C) 
D) 
E) 
In such a way, the largest volume corresponds to E) Silver, mass = 420 g, d = 10.5 g/cm.
Best regards.
Answer:
3.0×10⁻¹³ M
Explanation:
The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.
<h3>Equation</h3>
The power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.
Ksp = [Al₃⁺³]×[OH⁻]³
<h3>Solving for [Al₃⁺³]</h3>
The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.
This means the solubility product equation is ...
Ksp = x(10⁻⁷ +3x)³
For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...
3.0×10⁻³⁴ = x(10⁻⁷)³
Then x = [Al₃⁺³] will be ...
![[\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}](https://tex.z-dn.net/?f=%5B%5Ctext%7BAl%7D_3%5E%7B%5C%2C%2B3%7D%5D%3D%5Cdfrac%7B3.0%5Ctimes10%5E%7B-34%7D%7D%7B10%5E%7B-21%7D%7D%5Capprox%20%5Cboxed%7B3.0%5Ctimes10%5E%7B-13%7D%5Cqquad%5Ctext%7Bmoles%20per%20liter%7D%7D)
We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.
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<em>Additional comment</em>
The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.
Answer:
Normality of KOH = 0.1062 N
<em>Note: The given question is not clearly stated. A related question below is answered and explained;</em>
<em> What is the NORMALITY of a solution of KOH if 45.18 mL are required to neutralize 0.3g of pure oxalic acid H2C2O4 . 2H2O?</em>
Explanation:
Equation of the reaction: 2KOH + H₂C₂O₄.2H₂O --> K₂C₂O₄ + 4H₂O
molar mass 0f KOH = 56 g/mol; molar mass of H₂C₂O₄.2H₂O = 126 g/mol
Number of moles of in 0.3 g of H₂C₂O₄.2H₂O = mass/molar mass
number of moles = 0.3 g / 126 g/mol = 0.0024 moles
From the equation of reaction, 2 moles of KOH are required to neutralize 1 mole of H₂C₂O₄.2H₂O
Number of moles of KOH required to neutralize 0.0024 moles of H₂C₂O₄.2H₂O = 2 * 0.0024 = 0.0048 moles
Molar concentration of 45.18 mL solution of KOH containing 0.0048 moles of KOH = number of moles/volume (L)
molarity of KOH solution = (0.0048/45.18 mL) * (1000 mL/1 L) = 0.1062 mol/L
Normality = n * molarity
where is the number of moles of OH⁻ ions produces from mole of KOH = 1
Normality of KOH = 1 * 0.1062
Normality of KOH = 0.1062 N
Answer: This may result in the protein forming bonds with additional amino acids, this may result in the protein forming bonds with lipids instead of other proteins
Explanation:..