20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO =
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO
I think it turns into an igneous rock
In the modern periodic table the elements are arranged in order of increasing atomic number.
Answer:
ion
However atoms may gain or lose electrons in ordinary chemical reactions. If an atom has the same number of electrons as protons, it is a neutral atom. If it has a net charge, (more or less electrons than protons) it is an ion. If it has more electrons than protons it has a net negative charge and is known as an anion.
Explanation:
Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7