The approximate alcohol content is 210 ml.
Explanation:
It can be deduced from the question that each bottle is of 1000ml or 1 litre.
The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is
20/100*500
=100 ml
The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml
so it is 200ml having 30% alcohol
30/100*200
= 60 ml
The third bottle is one tenth full so its volume is 1/10*1000
100 ml. having 50% of alcohol
50/100*100
50 ml.
The alcohol content obtained from all these 3 litres is:
100+60+50
= 210 ml of alchohol is obtained from 800 ml of mixture.
Radio - Radio station transmits radio wavelength which is received by the
Radio.
<span>
Microwaves - Microwave Oven to heat up foods. </span>
<span>IR (infrared) - TV remote Control, to transmit IR light to a sensor in the TV so it can do some functions like increasing the volume, changing the channel etc. </span>
<span>Visible light - Sunlight or Light Bulbs </span>
<span>Ultraviolet - UV Lamps for sun tan, for detecting forged money </span>
<span>X-rays - Chest X-ray machines, Backscatter Xray (body scanner in airport security)
</span>
Gamma rays - Gamma rays<span> Medical Equipment which are used to kill cancer cells, to sterilize medical </span>equipment<span> </span>
"You may find what you seek, or you might not" Is an excellent quote that might help.
Answer:
Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Explanation:
The equilibriums that take place are:
Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.
For Cu⁺²:
2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
2.2x10⁻²⁰ = 0.25 M*[OH⁻]²
[OH⁻] = 2.97x10⁻¹⁰ M
For Co⁺²:
1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²
[OH⁻] = 7.21x10⁻⁸ M
<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Did you intend to write [PdCl4]^-2 instead of PdCl2-4? If so, then:
<span>Cathode: [PdCl4]^-2(aq) + 2e- ======⇒ Pd(s) + 4Cl-(aq) </span>
<span>Anode: Cd(s) ==⇒ Cd+2(aq) + 2e-</span>
