Yes Compounds are generally made of two or more atom or ion or species!
TRUE
Answer:
The value is ![U =1.3908 *10^{11} \ kg](https://tex.z-dn.net/?f=U%20%20%3D1.3908%20%2A10%5E%7B11%7D%20%5C%20%20kg%20)
Explanation:
From the question we are told that
mass of carbon dioxide produced by one gallon of gasoline is ![m = 9.50 \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%209.50%20%5C%20%20kg)
The number of cars is ![N = 40 \ million = 40 *10^{6} \ cars](https://tex.z-dn.net/?f=N%20%3D%20%2040%20%5C%20%20million%20%3D%20%2040%20%2A10%5E%7B6%7D%20%5C%20%20cars)
The distance covered by each car is ![d = 7930 \ mi](https://tex.z-dn.net/?f=d%20%3D%20%207930%20%5C%20mi)
The rate is ![R = 23.6 \ mi/ gallon](https://tex.z-dn.net/?f=R%20%20%3D%20%2023.6%20%5C%20%20mi%2F%20gallon)
Generally the amount of gasoline used by one car is mathematically represented as
![G = \frac{d}{R}](https://tex.z-dn.net/?f=G%20%20%3D%20%20%5Cfrac%7Bd%7D%7BR%7D)
=> ![G = \frac{7930}{23.6}](https://tex.z-dn.net/?f=G%20%20%3D%20%20%5Cfrac%7B7930%7D%7B23.6%7D)
=> ![G = 366 \ gallons](https://tex.z-dn.net/?f=G%20%20%3D%20%20%20366%20%5C%20%20gallons%20)
Generally the amount of gasoline used by N cars is
![H = N * G](https://tex.z-dn.net/?f=H%20%20%3D%20%20N%20%20%2A%20%20G)
=> ![H = 40*10^{6} * 366](https://tex.z-dn.net/?f=H%20%20%3D%20%2040%2A10%5E%7B6%7D%20%20%2A%20%20366)
=> ![H = 1.464*10^{10} \ gallons](https://tex.z-dn.net/?f=H%20%20%3D%201.464%2A10%5E%7B10%7D%20%5C%20%20gallons%20)
Generally the annual production of carbon dioxide is mathematically represented as
![U = m * H](https://tex.z-dn.net/?f=U%20%20%3D%20%20m%20%2A%20%20H)
=> ![U =9.50 * 1.464*10^{10}](https://tex.z-dn.net/?f=U%20%20%3D9.50%20%2A%20%20%201.464%2A10%5E%7B10%7D)
=> ![U =1.3908 *10^{11} \ kg](https://tex.z-dn.net/?f=U%20%20%3D1.3908%20%2A10%5E%7B11%7D%20%5C%20%20kg%20)
If you mean how many electrons an atom contains then this is the same as the atomic number of the atom. For example, the atomic number of oxygen is 8 so it contains 8 protons. Since in a stable atom, the number of protons is equal to the number of electrons, oxygen contains 8 electrons.
If you have an ion and you want to know how many electrons it has lost or gained, you look at the charge. If it is a 2+ ion then it has lost two electrons, if it's a - ion then it has gained an electron. The number of electrons in an ion is the atomic number +/- the charge shown. For example a Mg2+ ion would have 10 electrons as a magnesium atom has 12 electrons and loses two to form the 2+ ion.
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
![K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7BK_%7B%5Ctext%7Bw%7D%7D%7D%7BK_%7B%5Ctext%7Ba%7D%7D%7D%20%3D%20%5Cdfrac%7B1.00%20%5Ctimes%2010%5E%7B-14%7D%7D%7B2%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%205%20%5Ctimes%2010%5E%7B-10%7D)
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.