Answer: The speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>
Explanation:
Given, The speed of a race car = 99 miles/ hour
To convert the speed into kilometers per hour and kilometers per minute
Since 1 mile = 1.61 kilometers
So, Speed of car = (99 ) x (1.61 )
= 159.39 kilometers per hour.
Also, 1 hour = 60 minutes
Then, Speed of car = (159.39) ÷60
= 2.6565≈2.65 kilometer per minute.
Hence, the speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
Well it’s the first answer couches
The answer should be Pure substance
Answer:
The four resonance structures of the phenoxide ion are shown in the image attached
The conjugate base of cyclohexanol has only one resonance contributor, while
the conjugate base of phenol has four resonance contributors.
Explanation:
In organic chemistry, it is known that structures are more stable if they possess more resonance contributors. The greater the number of contributing canonical structures, the more stable the organic specie. Since the phenoxide ion has four contributing canonical structures, it is quite much more stable than cyclohexanol having only one contributing structure to its conjugate base. Hence the PKa(acid dissociation constant) of phenol is lesser than that of cyclohexanol. The conjugate base of phenol is stabilized by resonance.
