Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
Answer:
pH = 12.22
Explanation:
<em>... To make up 170mL of solution... The temperature is 25°C...</em>
<em />
The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:
Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)
<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>
<em />
To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:
<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>
0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =
2.80x10⁻³ moles of OH⁻
<em>Molarity [OH⁻] and [H⁺]</em>
2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M
As Kw at 25°C is 1x10⁻¹⁴:
Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]
[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M
<em>pH:</em>
pH = -log [H⁺]
pH = -log [6.068x10⁻¹³M]
<h3>pH = 12.22</h3>
Answer:
He had the most potential energy going down hill
Explanation:
because he picked up he went faster
The existence of spectra can help the existence of atoms by expanding and multiplying in the law of London Dispersion
