Answer:
sun rises in the east so to the right of the front of your house and sets to the left which is south.
Explanation: what i just said
Mass of CO₂ = 132 g
<h3>Further explanation
</h3>
A mole is a number of particles(atoms, molecules, ions) in a substance
This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mole also can be formulated :

moles of CO₂ = 3
mass of CO₂(MW=44.01 g/mol) :

Answer:
5.6 × 10^4
this is ur answer hope it helps u
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”.
<span>Assume
p=735 Torr
V= 7.6L
R=62.4
T= 295
PV-nRT
(735 Torr)(7.60L)= n (62.4Torr-Litres/mole-K)(295K)
0.30346 moles of NH3
Find moles
0.300L solution of 0.300 M HCL = 0.120 moles of HCL
0.30346 moles of NH3 reacts with 0.120 moles of HCL producing 0.120 moles of NH4+ ION, and leaving 0.18346 mole sof NH3 behind
Find molarity
0.120 moles of NH4+/0.300L = 0.400 M NH4+
0.18346 moles of NH3/0.300L = 0.6115 M NH3
NH4OH --> NH4 & OH-
Kb = [NH4+][OH]/[NH4OH]
1.8 e-5=[0.300][OH-]/[0.6115]
[OH-]=1.6e-5
pOH= 4.79
PH=9.21
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