Answer:
E = 19.89 × 10⁻¹³ J
Explanation:
Given data:
Wavelength of gamma ray= 1.0× 10⁻¹³ m
Energy of gamma ray = ?
Solution:
Formula:
E = h c/λ
h = plancks constant = 6.63× 10⁻³⁴ js
c = speed of light = 3 ×10⁸ m/s
by putting values,
E = 6.63× 10⁻³⁴ js × 3 ×10⁸ m/s /1.0× 10⁻¹³ m
E = 19.89 × 10⁻²⁶ J.m/ 1.0× 10⁻¹³ m
E = 19.89 × 10⁻¹³ J
A mixture in which two or more substances are evenly mixed, but not bonded together is a(n) homogeneous mixture, also called a(n) solution.
sorry if i got it wrong
Answer: There are several ways. The first that comes to mind is a pH meter. A pH electrode Is lowered into the solution, and (Assuming) the pH Meter has been properly calibrated, and the temperature of the solution is set to the calibration of the Meter, the pH can be read directly from an analogue scale or digital readout. Below 7 is acidic, 7 is Neutral, (like Pure Water), and over 7 is Alkaline, or Basic.
A useful, but less accurate method is the use of any number of “pH Indicator Solutions”, which are essentially a type of various colored dyes that change color within differing pH ranges. Usually, if the pH is unknown, a small amount of solution is removed from the container and tested separately - in a “well plate”, or similar method.
These types of dyes, or Indicator Solutions, can be dried upon strips of “pH indicator Paper”, which, depending upon the type can be very useful when carrying out more precisely arrived at pH tests like Titration.
Just to see if a solution is “Acid” or “Base”, Litmus paper is used; “a Red color shows Acidity, and a Blue color, a Base”; ergo, “An Acid Solution will turn Litmus Paper, Red”.
Answer:A typical gasoline mixture contains about 150 different hydrocarbons, including butane, pentane, isopentane and the BTEX compounds (benzene, ethylbenzene, toluene, and xylenes). Gasoline also contains chemicals such as lubricants, anti-rust agents and anti-icing agents that are added to improve car performance.
First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
