Answer:
The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of 
as indicated on the spectrum. The reaction between the 
and the 
are both clear liquids that form the orange liquid product which creates the absorbance spectrum. Because the color of the solution is orange, it reflects this and similar colors while absorbing blueish hues. We can find the absorption of only the by pre-rinsing the cuvette with each solution we intend to measure before placing it in the spectrometer. Also, wipe each cuvette with a kimwipe to remove all fingerprints that could effect the data collection.
Explanation:
The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of as indicated on the spectrum.
Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
Answer:
Solution of isopropanol is 10.25 molal
Explanation:
615 g of isopropanol (C3H7OH) per liter
We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.
Density of water 1g/mL → Density = Mass of water / 1000 mL of water
Notice we converted the L to mL
Mass of water = 1000 g (which is the same to say 1kg)
Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles
Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m
The advantage of having large vertebrae at the base of the vertebral column is having stability in terms of the center of gravity of the animal. If the animal has a large vertebrae, then it has an excellent balance and strength.
Answer : Both solutions contain 
molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain 
molecules.
Avogadro's Number is 
= 
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules = 
;
∴ Number of molecules = 
which will be = 
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
