<h3>Answer:</h3>
Platoic Acid
<h3>Explanation:</h3>
While naming Carboxylic Acids we know that when the Carboxylic Acid looses proton it is converted into corresponding conjugate base called as Carboxylate.
Examples:
HCOOH → HCOO⁻ + H⁺
Formic acid Formate Ion
H₃CCOOH → H₃CCOO⁻ + H⁺
Acetic acid Acetate Ion
H₅C₂COOH → H₅C₂COO⁻ + H⁺
Propanoic acid Propanoate Ion
Therefore, if the conjugate base is Platoate then the corresponding acid will be Platoic Acid means we will replace the -ate by -ic acid <em>i.e.</em>
RCOO⁻ + H⁺ → RCOOH
Platoate Ion Platoic Acid
While I am not the brainliest I can certainly answer.
This was a chemical change because the chemical components were changed, a big giveaway to this was the fizzing, however the temperature rising was also another giveaway.
In given data:
maximum absorption wavelength λ = 580 nm = 580 x 10⁻⁹ m
write the equation to find the crystal field splitting energy:
E = hC / λ
Here, E is the crystal field splitting energy, h = 6.63 x 10⁻³⁴ J.sec is Planck's constant and C = 3 x 10⁸ m/sec is speed of light.
substitute in the equation above:
E = (6.64 x 10⁻³⁴ x 3 x 10⁸) / (580 x 10⁻⁹) = 3.43 x 10⁻¹⁹J
This crystal field splitting energy is for 1 ion.
Number of atoms in one mole, NA = 6.023 x 10²³
to calculate the crystal field splitting energy for one mole:
E(total) = E x NA
= (3.43 x 10⁻¹⁹) x (6.023 x 10²³) = 206 kJ/ mole
The equilibrium constant is 0.0022.
Explanation:
The values given in the problem is
ΔG° = 1.22 ×10⁵ J/mol
T = 2400 K.
R = 8.314 J mol⁻¹ K⁻¹
The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.
The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K
So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.
We get,
So, the equilibrium constant is 0.0022.
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
