Answer: the basic difference is Exergonic reactions release energy and an endergonic reactions absorb energy .
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Answer:
Explanation:
Firstly, it should be noted that atomic number (number of protons) determines element. And the element with the atomic number 10 (10 protons) is Neon. Hence, Neon-10 (₁₀Ne) is the answer.
Note that sodium has an atomic number of 11. Also, number of protons is usually equal to the number of electrons in neutral atoms, this is because the total number of positive particles (protons) must be equal to the total number of negative particles (electrons) to give a neutral atom.
Answer:
KOH(aq) + HCI(aq) -----> KCI(aq )+ H2O
base acid salt water
hope this helps :)
Explanation:
Answer: The empirical formula of the compound becomes ![CH_2O](https://tex.z-dn.net/?f=CH_2O)
<u>Explanation:</u>
The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.
We are given:
Mass of C = 48.38 g
Mass of H = 6.74 g
Mass of O = 53.5 g
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
......(1)
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Putting values in equation 1, we get:
![\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20C%7D%3D%5Cfrac%7B48.38g%7D%7B12g%2Fmol%7D%3D3.023%20mol)
![\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20H%7D%3D%5Cfrac%7B6.74g%7D%7B1g%2Fmol%7D%3D6.74%20mol)
![\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20O%7D%3D%5Cfrac%7B53.5g%7D%7B1g%2Fmol%7D%3D3.34%20mol)
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles
![\text{Mole fraction of C}=\frac{3.023}{3.023}=1](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20C%7D%3D%5Cfrac%7B3.023%7D%7B3.023%7D%3D1)
![\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20H%7D%3D%5Cfrac%7B6.74%7D%7B3.023%7D%3D2.23%5Capprox%202)
![\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1](https://tex.z-dn.net/?f=%5Ctext%7BMole%20fraction%20of%20O%7D%3D%5Cfrac%7B3.34%7D%7B3.023%7D%3D1.105%5Capprox%201)
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1 : 2 : 1
Hence, the empirical formula of the compound becomes ![CH_2O](https://tex.z-dn.net/?f=CH_2O)