You can use the formula M1 x V1 = M2 x V2 where M1 is the molarity of the first substance and V1 is the volume of the substance. M2 is the molarity of the 2nd substance and V2 is the volume of the substance
if substance 1 is HCl and 2 is KOH we can set up the following equation
x moles / liter (unknown) x .02 liters = .5 moles/ liter x .032 liters
x moles / liter x .02 liters = .016 moles
x moles / liter = .016/.02 liters
x moles / liter = .8 moles/ liter
You have . 8M HCL
Gold potential is the correct answer
Answer:
22m/s
Explanation:
Given parameters:
Time = 5s
Acceleration = 4m/s²
Initial velocity = 2m/s
Unknown
Final speed = ?
Solution:
To solve this problem, we use the expression below;
v = u + at
v is the final speed
u is the initial speed
a is the acceleration
t is the time taken
So, insert the parameters and solve;
v = 2 + 4 x 5 = 22m/s
Answer:
One complete revolution around a circular path.
Explanation:
Let us take the case of a car moving in a circular track of radius r metres.
In one revolution, the car covers the length(distance) equal to the perimeter of the circle.
In this case, distance traveled = 2
r metres
But after one complete revolution, the car reaches the same position as it was at the beginning of the motion.
Hence, the initial and final points coincide or the car hasn't changed it's position w.r.t the initial point.
So in this case, the displacement is zero.
Hence, revolution of a car around a circular path is an example of an object traveling a distance but having no displacement.
Answer:
Ammonia > Urea > Ammonium nitrate > Ammonium sulphate
Explanation:
Percentage by mass of nitrogen in NH3:
Molar mass of NH3= 17 g/mol
Hence % by mass = 14/17 × 100 = 82.35%
% by mass of NH4NO3
Molar mass of NH4NO3 = 80.043 g/mol
Hence; 28/80.043 × 100 = 34.98%
% by mass of (NH4)2SO4;
Molar mass of (NH4)2SO4= 132.14 g/mol
Hence; 28/132.14 × 100 = 21.19%
% by mass of CH4N2O
Molar mass of urea = 60.0553 g/mol
Hence 28/60.0553 × 100 = 46.62%
