Rows are periods, columns are groups
Answer:
No net change in reaction occurs in this nucleophilic acyl subtitution reaction
Explanation:
Sodium ethoxide in ethanol gives nucleophilic acyl substitution reaction with ethyl-2-methylpropanoate.
Here ethoxide group replaces an ethoxide group from ester through addition-ellimination pathway.
So, ultimately, the product of this reaction is identical with reactant i.e. ethyl-2-methylpropanoate is reproduced.
Hence one might observe no change during reaction as product and reactant of this reaction are same.
Mechanistic pathway has been shown below.
The answer is Rubidium (Rb)
Answer:
1.552 moles
Explanation:
First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.
This is illustrated below:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 = 22.1g
Mole of C8H18 =..?
Number of mole = Mass /Molar Mass
Mole of C8H18 = 22.1/144
Mole of C8H18 = 0.194 mole
From the balanced equation above,
2 moles of C8H18 produced 16 moles of CO2.
Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.
Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned
Answer:
157.64 L
Explanation:
We'll begin by converting 30 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 30 °C
T(K) = 30 °C + 273
T (K) = 303 K
Next, we shall convert 600 mmHg to atm. This can be obtained as follow:
760 mmHg = 1 atm
Therefore,
600 mmHg = 600 mmHg × 1 atm / 760 mmHg
600 mmHg = 0.789 atm
Finally, we shall determine the volume of the gas. This can be obtained as follow:
Number of mole (n) = 5 moles
Temperature (T) = 303 K
Pressure (P) = 0.789 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
PV = nRT
0.789 × V = 5 × 0.0821 × 303
0.789 × V = 124.3815
Divide both side by 0.789
V = 124.3815 / 0.789
V = 157.64 L
Therefore, the volume of the gas is 157.64 L
