HALOGENS have 7 electrons in their final shell regardless if the total number of atoms. These 7 electrons would be shared between the 's' and 'p' orbital in the form
ns² np⁵ (the same as the question). Attached is a table that highlights the fact that halogens have the same form in terms of the valence electrons.
Solution: (a) The net force is zero because car is travelling with same speed that is 60 miles per hour or there is no acceleration.
(b) Mass of object is 10 kg
Acceleration is 2 m/s², thus force can be calculated as follows:
F=m×a
Here, m is mass of object and a is its acceleration thus,
F=10 kg×2 m/s²
=20 kg m/s²
Here, 1 N=1 kg m/s²
Thus, F=20 N
(c) Force on object is 17 N and its acceleration is 1.5 m/s².
Mass of the object can be calculated as follows:
m=\frac{F}{a}
thus,
m=\frac{17 N}{1.5 m/s^{2}}=11.33 (\frac{N}{m/s^{2}})
Here, 1 N=1 kg m/s²
Thus,
m=11.33 kg
Answer:
-476.95 Kj
Explanation:
N2H4(l) + N2O4(g) = 2N2O(g) + 2H20(g)
∆Hrxn = n∆Hf(products) - m∆Hf(reactants)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation, ∆Hf = standard enthalpy of formation, ∆Hrxn= standard enthalpy of reaction.
Using the following standard enthalpies of formation ( you did not provide any ):
N2H4(l) = +50.63Kj/mol; N2O4(g) = +9.08Kj/mol; N2O(g) =+33.18Kj/mol; H2O(g) = -241.8Kj/mol
∆Hrxn = [ (2(∆Hf(N2O)) + (2(∆Hf(H2O))] – [(1(∆Hf(N2H4)) + (1(∆Hf(N2O4))]
∆Hrxn = [ 2(+33.18) + 2(-241.8)] – [ (+50.63) + (+9.08)]
∆Hrxn = [ (+66.36)+(-483.6)] – [ +50.63+9.08]
∆Hrxn = [ +66.36-483.6] – [+59.71]
∆Hrxn = -417.24-59.71
∆Hrxn = -476.95 Kj
NOTE: Remember to use the standard enthalpies of formation given to you by your instructor if they differ from the values used herein, and follow the same procedure.
Answer:
The correct equation will be "[Fe(H2O)5]²⁺ + NO → Fe(H2O)5NO]²+ (brown ring)
".
Explanation:
For NO₂, the addition of H₂SO₄ forms HNO₂ comprising NO gas as well as HNO₂, the equation will be:
⇒ NO₂⁻ + 2H⁺ → HNO2
⇒ HNO₂ + H₂O + H⁺ → HNO₃ + NO
For NO₃⁻,
NO₃⁻, Fe₂+ oxidized to Fe₃ + then releases NO gas in the existence of H₂SO₄
⇒ 3Fe₂⁺ + 4H⁺ + NO₃⁻ → 3Fe₃⁺ + NO + 2H₂O
Brown ring forming establishes NO3- presence in the initial test sample,
⇒ [Fe(H2O)5]²⁺ + NO → Fe(H2O)5NO]²+ (brown ring)
The increase in the average kinetic energy of the molecules needs the increase of temperature. The relation of ℃ and K is n ℃ = m K -273. So if we change all the K unit to ℃ unit, we can get the answer is (1).
