Answer:
The final volume will be 24.7 cm³
Explanation:
<u>Step 1:</u> Data given:
Initial temperature = 180 °C
initial volume = 13 cm³ = 13 mL
The mixture is heated to a fina,l temperature of 587 °C
Pressure and amount = constant
<u>Step 2: </u>Calculate final volume
V1/T1 = V2/T2
with V1 = the initial volume V1 = 13 mL = 13*10^-3
with T1 = the initial temperature = 180 °C = 453 Kelvin
with V2 = the final volume = TO BE DETERMINED
with T2 = the final temperature = 587 °C = 860 Kelvin
V2 = (V1*T2)/T1
V2 = (13 mL *860 Kelvin) /453 Kelvin
V2 = 24.68 mL = 24.7 cm³
The final volume will be 24.7 cm³
Answer:
Explanation:
1) Vanadium (IV) → V⁺⁴
Carbonate → CO₃⁻²
So , Vanadium (IV) Carbonate = V₂(CO₃)₄ or V(CO₃)₂
2) Tin (II) = Sn⁺²
Nitrite = NO₂⁻
So, Tin (II) Nitrate = Sn(NO₂)₂
3) Cobalt (III) = Co⁺³
Oxide = O⁻²
So , Cobalt (III) Oxide = Co₂O₃
4) Titanium (II) = Tn⁺²
Acetate = CH₃COO⁻
So , Titanium (II) Acetate = Tn(CH₃COO)₂ or Tn(C₂H₃O₂)₂
5) Vanadium (V) = V⁺⁵
Sulfide = S⁻²
So , Vanadium (V) Sulfide = V₂S₅
6) Chromium (III) = Cr⁺³
Hydroxide = OH⁻
So , Chromium (III) Hydroxide = Cr(OH)₃
7) Lithium = Li⁺
Iodide = I⁻
So , Lithium Iodide = LiI
8) Lead (II) = Pb⁺²
Nitride = N⁻³
So , Lead (II) Nitride = Pb₃N₂
9) Silver = Ag⁺
Bromide = Br⁻
So , Silver Bromide = AgBr
It means the relationship between two systems connected only by a diathermic wall.
Consider the acid spill. It is already starting to do nasty things to, say, the floor or counter. So you grab the bottle of 10% NaOH and pour some on the spill. All of a sudden, you get a great deal of heat, and you don't have any visual evidence whether your put on too little or too much. But you have added more liquid to the spill, generated more heat, and will get more damage. You have made a bigger mess, and if you added too much, you then have a neutralization problem to deal with.
And if it is something like a strong sulfuric acid solution, adding sodium hydroxide solution will be extremely exothermic, and you could get some really nasty results.
So now approach the spill with a handful of baking soda. You sprinkle it on the spill. It fizzes, and carbon dioxide is given off. That actually, in a very tiny way, moderates the temperature of the neutralization. And you can keep adding baking soda until the fizzing stops, and then perhaps some water to mix everything well. But what you have done is kept the volume to a minimum, added a neutralization agent that has a visible endpoint (no more gas being given off), and you don't suddenly have a huge amount of highly basic solution because you added too much.
And what is also nice about baking soda is that you can toss some with your hand or even with a spoon, and get some distance from the spill. With a liquid, you have to get much closer
i hope this helped..
