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Answer:
A homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions. A heterogeneous mixture consists of visibly different substances or phases. The three phases or states of matter are gas, liquid, and solid.
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:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
0.256 L
Explanation:
We should use the following formula:
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.82 M NaOCl
volume (1) = ?
concentration (2) = 0.21 M NaOCl
volume (2) = 1 L
volume (1) = [concentration (2) × volume (2)] / concentration (1)
volume (1) = [0.21 / 1] / 0.82 = 0.256 L
