B and D is out. It cant be A because heat of combustion is substance not compound. So the answer is D.
Answer is: a) I only.
Above critical temperature of CO₂, a gas cannot be liquefied no matter how much pressure is applied. Temperature and pressure above its critical point is called supercritical fluid and this is <span>intermediate between gaseous and liquid states.</span>
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Answer:
it's
C. with melting point 11° C and boiling point 181° C
Explanation:
i hope you got it


- <u>We </u><u>have </u><u>250g </u><u>of </u><u>liquid </u><u>water </u><u>and </u><u>it </u><u>needs </u><u>to </u><u>be </u><u>cool </u><u>at </u><u>temperature </u><u>from </u><u>1</u><u>0</u><u>0</u><u>°</u><u> </u><u>C </u><u>to </u><u>0</u><u>°</u><u> </u><u>C</u>
- <u>Specific </u><u>heat </u><u>of </u><u>water </u><u>is </u><u>4</u><u>.</u><u>1</u><u>8</u><u>0</u><u>J</u><u>/</u><u>g</u><u>°</u><u>C</u>

- <u>We </u><u>have </u><u>to </u><u>find </u><u>the</u><u> </u><u>total</u><u> </u><u>number </u><u>of </u><u>joules </u><u>released</u><u>. </u>

<u>We </u><u>know </u><u>that</u><u>, </u>
Amount of heat energy = mass * specific heat * change in temperature
<u>That </u><u>is, </u>

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>




Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

<u>We </u><u>have </u><u>to </u><u>tell </u><u>whether </u><u>the </u><u>above </u><u>process </u><u>is </u><u>endothermic </u><u>or </u><u>exothermic </u><u>:</u><u>-</u>
Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic
- <u>Exothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>evolved </u><u>. </u>
- <u>Endothermic </u><u>process </u><u>:</u><u>-</u><u> </u><u>It </u><u>is </u><u>the </u><u>process </u><u>in </u><u>which </u><u>heat </u><u>is </u><u>absorbed </u><u>.</u>