Answer:
Given, Mass of an object, m = 2kg
velocity, v = 2m/s
We have,
KE = 1/2 m v²
= 1/2 × 2 (2)²
= 1/2 × 8
= 4J
Hence, 4J is the required Kinetic energy.
Answer:
c-do background research.
Explanation:
The scientific methods exposes the way scientists carry out their investigation and how findings are reported and at times discarded. It is systematic way of studying perceived observations in the environment.
During the course of an investigation, scientists may seek to find out about the prevailing knowledge about a phenomena and the level of research that might have been done in that regard. It is proper for such a scientist to do a background research by searching for related journals and publications in that field.
Answer:
53.11× 10²³ molecules
Explanation:
Given data:
Number of molecules of CO₂ = ?
Mass of CO₂ = 388.1 g
Solution:
Formula:
Number of moles = mass/ molar mass
Molar mass of CO₂ = 12× 1 + 16×2
Molar mass of CO₂ = 44 g/mol
Now we will put the values in formula.
Number of moles = 388.1 g/ 44 g/mol
Number of moles = 8.82 moles
Now we will calculate the number of molecules by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
8.82 mol × 6.022 × 10²³ molecules / 1 mol
53.11× 10²³ molecules
Answer:
58.9mL
Explanation:
Given parameters:
Initial volume = 34.3mL = 0.0343dm³
Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³
Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³
Unknown:
Final volume =?
Solution:
Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.
Therefore;
C₁V₁ = C₂V₂
where C and V are concentration and 1 and 2 are initial and final states.
now input the variables;
1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂
V₂ = 0.0589dm³ = 58.9mL
