Answer: A
Explanation: The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.
These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.
Answer:
4.002 mol is in 68.16 grams of NH3
Answer:
Limiting reagent: AgNO3
grams AgCl : 2.44 g AgCl
grams of excess reagent remain: 0.62 g BaCl2
Explanation:
1. Change grams to mol:
<em>AgNO3:</em>
5.738g x (1mol/169.87g) = 0.034 mol AgNO3
<em>BaCl2:</em>
4.115g x (1 mol/208.23g) = 0.020 mol BaCl2
2. Limiting reagent:
<em>AgNO3:</em>
0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2
<em>BaCl2:</em>
0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3
Limiting reagent: AgNO3
3. Grams of AgCl produced:
Using the limiting reagent:
0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl
4. Change mol to grams:
0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl
5. Grams of the excess reagent:
0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2
0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2
0.003 mol BaCl2 x ( 208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2