Answer:
C. 30 kJ
Explanation:
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In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:
It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:
Therefore, the answer is C. 30 kJ
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Answer: Potassium hydroxide, KOH, is considered a <u>BASE</u> in an acid-base reaction because it <u>ACCEPTS</u> a hydrogen ion from the other reactant.
According to Brønsted–Lowry acid–base theory, Base is a specie which accepts proton (H⁺) while, Acid is a specie which donate proton.
Bases may contain a negative charge or lone pair of electrons, while, Acids contain positive charge or a neutral atom with incomplete octet.
In given statement KOH is acting as a base because it contains a negatively charged hydroxyl group which can accept proton from a acid, i.e.
KOH → K⁺ + OH⁻
Reaction of OH⁻ with any acid,
K⁺ + OH⁻ + HCl → H₂O + KCl
The number of oxygen atoms is in 5.00 g of sample of sodium dichromate, Na₂Cr₂O₇ is 1.125 x 10²³ atoms.
<h3>What is sodium dichromate?</h3>
Sodium dichromate is an inorganic compound. It is sued in tanning and mental illness.
The molar mass of sodium = 23 gm
The molar mass of Cr = 51.99 gm
The molar mass of oxygen = 16 gm
Molar mass of Na₂Cr₂O₇ = 2(23) + 2(51.99) + 7(16) = 261.98 gm
Number of moles in 7 gm = mass / molar mass = 7 / 261.98 = 0.0267 moles
1 mole of Na₂Cr₂O₇ contains 7 moles of oxygen, therefore:
The number of moles of oxygen in 0.0267 moles = 0.0267 x 7 = 0.1869 moles
Thus, the number of atoms = 0.1869 x 6.02 x 10^23 = 1.125 x 10²³ atoms.
To learn more about sodium dichromate, refer to the link:
brainly.com/question/1444529
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<span>The arrangement of the periodic table leads us to visualize certain trends among the atoms. The vertical columns ( basically their groups) of the periodic table and they are arranged such that all its elements have the same number of valence electrons. And all the elements within a certain group thus share similar properties and that's about it.</span>
Im sorry , Im confused reading this . maybe enter a image ?
