What is research? Depending on who you ask, you will likely get very different answers to this seemingly innocuous question. Some people will say that they routinely research different online websites to find the best place to buy goods or services they want. Television news channels supposedly conduct research in the form of viewer polls on topics of public interest such as forthcoming elections or government-funded projects. Undergraduate students research the Internet to find the information they need to complete assigned projects or term papers. Graduate students working on research projects for a professor may see research as collecting or analyzing data related to their project. Businesses and consultants research different potential solutions to remedy organizational problems such as a supply chain bottleneck or to identify customer purchase patterns. However, none of the above can be considered “scientific research” unless: (1) it contributes to a body of science, and (2) it follows the scientific method. This chapter will examine what these terms mean
pH of buffer can be calculated as:
pH=pKa+log[salt]/[Acid]
As ka = 4.58 x 10-4
Concentration of [Salt] that is NO2(-1)=0.380M
Concentration of [Acid] that is HNO2=0.500M
So, pH= -log(4.58*10^-4)+log((0.380)/0.500))
=3.21
So pH of solution will be 3.21
Answer:
Newton's laws are very important because they tie into almost everything we see in everyday life. These laws tell us exactly how things move or sit still, like why you don't float out of bed or fall through the floor of your house.
Explanation: Newton's laws of motion are important because they are the foundation of classical mechanics, one of the main branches of physics. Mechanics is the study of how objects move or do not move when forces act upon them.
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
