Answer:
1.024 x 10⁻²²g
Explanation:
Data Given:
mass of copper = 63.5 g
no. of atoms of copper = 6.02 x10²³ atoms
mass of of an average copper = ?
Solution:
As 6.02 x10²³ atoms have 63.5 g of mass then what will be the mass of atom.
Apply unity formula
63.5 g of copper ≅ 6.02 x10²³ atoms of copper
mass of copper atom ≅ 1 atom of copper
Do cross multiplication
mass of copper atom = 1 atom x 63.5 g / 6.02 x10²³ atoms
mass of copper atom = 1.024 x 10⁻²² g
mass of an average copper atom = 1.024 x 10⁻²² g
Generally, the top elements in group 16 will have a charge of -2, although the entire group isn't standardized.
Answer;
H- (negatively charged hydrogen ion)
Explanation;
The hydrogen ion has the same electron configuration as an atom of Helium. -Hydrogen ion is formed when a hydrogen atom loses or gains an electron. A positively charged ion is formed when hydrogen atom loses an electron (remaining with zero electrons) while a negatively charged ion (H-) is formed when a hydrogen atom gains an electron thus having 2 electrons.
-Helium atom has two electrons, therefore, it is similar with an negatively charged hydrogen.
Answer:
SiCl4 + H2O = SiO2 + HCl
Reactant. product
Si = 1 atom Si = 1 atom
Cl = 4 atom Cl = 1 atom
H = 2 atom H = 1 atom
O = 1 atom O = 2 atom
SiCl4. + 2H2o = SiO2 + 4hcl
Reactant. product
Si = 1 atom Si = 1 atom
Cl = 4 atom Cl = 4 atom
H = 4 atom H = 4 atom
O = 2 atom O = 2 atom
Balanced chemical equation.
Answer:
Here's what I get
Explanation:
1. Write the chemical equation
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵
Let's rewrite the equation as
A⁻ + H₂O ⇌ HA + OH⁻
2. Calculate Kb
3. Set up an ICE table
A⁻ + H₂O ⇌ HA + OH⁻
I/mol·L⁻¹: 0.35 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.35-x x x
4. Solve for x
![\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BHA%20%5D%5BOH%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%20%3D%20%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.35-x%7D%20%3D%205%20%5Ctimes%2010%5E%7B-10%7D)
Check for negligibility,
![\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BHA%5D%7D%7D%7BK_%7B%5Ctext%7Bb%7D%7D%7D%20%3D%20%5Cdfrac%7B0.35%7D%7B5%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%207%20%5Ctimes%2010%5E%7B8%7D%3E%20400%5C%5C%5C%5C%5Ctherefore%20x%20%5Cll%200.35%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.35%7D%20%3D%205%20%5Ctimes%2010%5E%7B-10%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.35%20%5Ctimes%205%20%5Ctimes%2010%5E%7B-10%7D%20%3D%201.8%5Ctimes%2010%5E%7B-10%7D%5C%5C%5C%5Cx%20%3D%20%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1%20%5Ctimes%2010%5E%7B-5%7D%7D)
5. Calculate the pOH
[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹
pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88
6. Calculate the pH.
pH + pOH = 14.00
pH + 4.88 = 14.00
pH = 9.12
Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.
