Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
Answer : The correct answer is 74.83 m/s .
The kinetic energy is energy possessed by any mass which is moving or have some speed . It is product of mass and velocity . It is expressed as :

Where KE = kinetic energy in J or
m = mass in Kg v = speed in m/s²
Unit of KE is Joules (J) .
Givne : KE = 2100 J mass = 0.75 kg v = ?
Plugging value in KE formula =>


Dividing both side by 0.375 kg =>



Answer:
option "B" is correct (substance 2)
Is the answer youre looking for 377297 feet