The element of the group 17 that is most active non metal is fluorine.
The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).
Among all the elements of the group 17. Fluorine is the smallest in size.
Because of the small size of fluorine it has the highest electronegativity in group 17.
This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.
Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.
It reacts readily to form oxides and hydroxides.
So, we can conclude here that fluorine is the most active non metal of group 17.
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Answer:
n ≈ 2.42 moles
General Formulas and Concepts:
<u>Chem</u>
Ideal Gas Law: PV = nRT
- P is pressure in mmHg
- V is volume in liters
- n is number of moles
- R is a constant (62.4 L · mmHg/mol · k)
- T is temperature in Kelvins
K = °C + 273
Explanation:
<u>Step 1: Define</u>
V = 50.2 L
P = 755 mmHg
T = -22.0°C = 251 K
<u>Step 2: Find moles </u><em><u>n</u></em>
(755 mmHg)(50.2 L) = n(62.4 L · mmHg/mol · k)(251 K)
37901 mmHg · L = n(15662.4 L · mmHg/mol)
n = 2.41987 moles
<u>Step 3: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules.</em>
2.41987 moles ≈ 2.42 moles
The answer is ultraviolet
Answer:
Solar Power, HEP (Hydroelectric Power), Wind Power, etc...
Explanation:
These are some common examples of renewable energy source.
Answer:
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Explanation:
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).
Silver chromate is the salt of a strong base (AgOH) and a weak acid (H₂CrO₄).
HCrO₄⁻ is an even weaker acid than H₂CrO₄, so CrO₄²⁻ is a strong base.
Any added H⁺ will immediately combine with the chromate ions according to the reaction
H⁺ + CrO₄²⁻ ⟶ HCrO₄⁻
thereby removing chromate ions from solution.
According to Le Châtelier's Principle, more silver chromate will dissolve to replace the chromate ions that the H⁺ removes.
The overall equation for the reaction is
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + <em>CrO₄²⁻(aq)
</em>
<u>H⁺(aq) + </u><em><u>CrO₄²⁻(aq)</u></em><u> ⟶ HCrO₄⁻(aq)
</u>
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)