Answer : The correct option is A.
Explanation :
Law of conservation of mass : In the chemical reaction, the mass of reactant must be equal to the mass of product.
A. 
The mass of reactant side = ![2C_4H_{10}+2Cl_2+12O_2=[8(12)+20(1)]+4(35.5)+24(16)=642g](https://tex.z-dn.net/?f=2C_4H_%7B10%7D%2B2Cl_2%2B12O_2%3D%5B8%2812%29%2B20%281%29%5D%2B4%2835.5%29%2B24%2816%29%3D642g)
The mass of product side = 
This means, the mass of product is equal to the mass of reactant. The mass remains conserved and obeys the law of conservation of mass.
The reaction B, C, D, E does not obey the law of conservation of mass.
Therefore, Only reaction A obey the law of conservation of mass.
Answer: He theorized that all material bodies are made up of indivisibly small “atoms.” Aristotle famously rejected atomism in On Generation and Corruption. Aristotle refused to believe that the whole of reality is reducible to a system of atoms.
Explanation:
Answer:
Tapeworm, also called cestode, any member of the invertebrate class Cestoda (phylum Platyhelminthes), a group of parasitic flatworms containing about 5,000 species. ... Tapeworms also lack a circulatory system and an organ specialized for gas exchange.
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.