Answer:
We need 5.55 mL of H2SO4
Explanation:
Step 1: Data given
96% (w/w) pure H2SO4
We want to prepare 500.00 mL of a 2.0 M solution of H2SO4
Molar mass of H2SO4 is 98.08 g/mol
Density of H2SO4 = 1.84 g/mL
Step 2: Calculate moles H2SO4
0.2 M means we have 0.2 moles / 1L
Moles = molarity * volume
Moles = 0.2 M * 0.500 L = 0.100 moles of H2SO4
0.2M = 0.2 mole/L
Therefore, 0.5L * 0.2M = 0.1 mole of H2SO4 required.
Step 3: Calculate mass H2SO4 required
Mass = moles * molar mass
Mass = 0.100 moles * 98.08 g/mol = 9.808 grams
Step 4: Calculate volume needed
Volume = mass / density
Volume = 9.808 grams / 1.84 g/mL
Volume = 5.33 mL
This would be if the purity of H2SO4 was 100%
The purity of your substance is 96%, therefore you will actually need:
Volume needed = 5.33*(1/0.96) = 5.55 mL.
We need 5.55 mL of H2SO4
Answer: The standard potential is -0.141 V
Explanation:
To calculate the Gibbs free energy for given value of equilibrium constant we use the relation:
![\Delta G=-RTlnK](https://tex.z-dn.net/?f=%5CDelta%20G%3D-RTlnK)
where,
= standard Gibbs free energy = ?
R = Gas constant = 8.314 J/Kmol
T = temperature = 298 K
K = equilibrium constant =
Putting values in above equation, we get:
![\Delta G=40853J](https://tex.z-dn.net/?f=%5CDelta%20G%3D40853J)
Also ![\Delta G=-nFE^0](https://tex.z-dn.net/?f=%5CDelta%20G%3D-nFE%5E0)
where n = no of electrons gained or lost = 3
F = Faradays constant = 96500 C
= standard potential = ?
![40853=3\times 96500\times E^0](https://tex.z-dn.net/?f=40853%3D3%5Ctimes%2096500%5Ctimes%20E%5E0)
![E^0=-0.141V](https://tex.z-dn.net/?f=E%5E0%3D-0.141V)
Thus the standard potential is -0.141 V
Hey there!
I assume you want this equation balanced.
Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄
Let's start by balancing the polyatomic ions, starting with SO₄.
There are 3 on the left and 1 on the right, so let's add a coefficient of 3 in front of the CaSO₄.
Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + 3CaSO₄
Now let's balance the polyatomic ion OH.
There are 2 on the left and 3 on the right, so let's put a coefficient of 3 in front of Ca(OH)₂ and a coefficient of 2 in front of Al(OH)₃.
Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄
Now let's balance Al.
There are 2 on the left and 2 on the right, so Al is already balanced.
Lastly, let's balance Ca.
There are 3 on the left and 3 on the right, so Ca is already balanced.
This means we're done! Here's the final balanced equation:
Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄
Hope this helps!
he value for the equilibrium constant for the following chemical reaction, the auto-ionization of water, is 1.0x10-14 at 298 K <u>1x10-14 0.5x10-14 2x10-14 -1x10-14 1x1014 1x10-15</u>
<h3>What is
chemical reaction?</h3>
A chemical reaction is a procedure that causes one group of chemical components to change chemically into another. Chemical reactions, which can frequently be described by a chemical equation, traditionally include changes that only affect the locations of electrons in the formation and dissolution of chemical bonds between atoms, with no change to the nuclei (no change to the elements present). The study of chemical processes involving unstable and radioactive elements, where both electronic and nuclear changes may take place, is known as nuclear chemistry.
Reactants or reagents are the substance(s) or substances that are initially utilized in a chemical reaction.
To learn more about chemical reaction, from the given link:
brainly.com/question/14197404
#SPJ4