Choose the isotope that would be most stable. carbon-14 barium-56 lead-82 radon-85

ddd [48]

Electromagnetic waves

7 0

3 years ago

HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make

vazorg [7]

Answer: The correct statement is low temperature only, because entropy decreases during freezing.

Explanation:

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = change in Gibb's free energy

$\Delta H$ = change in enthalpy

T = temperature

$\Delta S$ = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and $Delta S$ is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the $\Delta H$ is also negative.

For a reaction to be spontaneous, $\Delta G$ must be negative.

$-ve=-ve-[T(-ve)]\\\\-ve=-ve+T$

From above equations, it is visible that $\Delta G$ will be negative only when the temperature will be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

8 0

3 years ago

2. Why are stain used when preparing observing cells under the

ella [17]

Answer:

to see the cells clearly

8 0

3 years ago

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Calculate the volume of 150kg of chlorine gas at room temperature and pressure. The volume of one mole of gas at room temperatur

Tom [10]

Answer:

thats now easyyy....................❤.you have to be genuine

Explanation:

gghggg

plssss

6 0

3 years ago

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Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det

mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0

3 years ago