Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁴
(b) 1s² 2s² 2p⁶ 3s² 3p⁵
(c) sp³
(d) No valence orbital remains unhybridized.
Explanation:
<em>Consider the SCl₂ molecule. </em>
<em>(a) What is the electron configuration of an isolated S atom? </em>
S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.
<em>(b) What is the electron configuration of an isolated Cl atom? </em>
Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>
SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.
<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>
No valence orbital remains unhybridized.
a scale-model mound made of the same materials that make the real hill
Answer:
The answer to your question is 41.6 g of AgCl
Explanation:
Data
mass of NH₄Cl = 15.5 g
mass of AgNO₃ = excess
mass of AgCl = 35.5 g
theoretical yield = ?
Process
1.- Write the balanced chemical reaction.
NH₄Cl + AgNO₃ ⇒ AgCl + NH₄NO₃
2.- Calculate the molar mass of NH₄Cl and AgCl
NH₄Cl = 14 + 4 + 35.5 = 53.5 g
AgCl = 108 + 35.5 = 143.5 g
3.- Calculate the theoretical yield
53.5 g of NH₄Cl -------------------- 143.5 g of AgCl
15.5 g of NH₄Cl ------------------- x
x = (15.5 x 143.5) / 53.5
x = 2224.25 / 53.5
x = 41.6 g of AgCl
<span>We look at the end of the day:
n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
