Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
Answer:
- <u>Dependent variable</u>
Explanation:
In a controlled <em>experiment </em>there will be one independent variable, one dependent variable, and some controlled or constant parameters.
The target of the expermiment is to tes how the dependent variable changes with the independent variable.
So, the scientist will change (manipulate) the independent variable and measure the dependent vabiales.
Here, the temperature will be manipulated (within a range), so this is the independent variable; the height of the cake is the variable whose behavior wants to be determined, so this is the dependent variable. Controlled variables or constants may include the mixture, the size of the pan, the time of cooking, the oven, among others.
Answer:
194.6 mL of SO₂
Explanation:
The reaction that takes place is:
P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)
<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:
- 23.8 °C → 23.8 + 273.15 = 296.95 K
- 747 torr → 747/760 = 0.983 atm
We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):
0.576 g P₄S₃ * 
= 7.85 * 10⁻³ mol SO₂ = n
PV=nRT
0.983 atm * V = 7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K
V = 0.1946 L
- Finally we convert L into mL:
0.1946 * 1000 = 194.6 mL
<h2>It can happen when liquids are cold or when they are warm. ... It turns out that all liquids can evaporate at room temperature and normal air pressure. Evaporation happens when atoms or molecules escape from the liquid and turn into a vapor. Not all of the molecules in a liquid have the same energy.</h2>
The given question is incomplete. The complete question is as follows.
Sodium sulfate is slowly added to a solution containing 0.0500 M 
and 0.0390 M 
. What will be the concentration of 
(aq) when 
begins to precipitate? What percentage of the can be separated from the Ag(aq) by selective precipitation?
Explanation:
The given reaction is as follows.
![[Ag^{+}]](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D)
= 0.0390 M
When 
precipitates then expression for 
will be as follows.
![K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]](https://tex.z-dn.net/?f=1.20%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%280.0390%29%5E%7B2%7D%20%5Ctimes%20%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![[SO^{2-}_{4}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B4%7D%5D)
= 0.00788 M
Now, equation for dissociation of calcium sulfate is as follows.
![K_{sp} = [Ca^{2+}][SO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BSO%5E%7B2-%7D_%7B4%7D%5D)
![4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788](https://tex.z-dn.net/?f=4.93%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5BCa%5E%7B2%2B%7D%5D%20%5Ctimes%200.00788)
![[Ca^{2+}]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D)
= 0.00625 M
Now, we will calculate the percentage of remaining in the solution as follows.
= 12.5%
And, the percentage of that can be separated is as follows.
100 - 12.5
= 87.5%
Thus, we can conclude that 87.5% will be the concentration of when begins to precipitate.
