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butalik [34]
2 years ago
10

How does an ice cube particles turn to a gas by laying in the sun?

Chemistry
2 answers:
natali 33 [55]2 years ago
7 0
It Evaporators after being in the sun
denpristay [2]2 years ago
5 0
It evaporates so the ice cube turns in to water then after words evaporat
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Determine the correct set of coefficients to balance the chemical equation. __c6h6(l) + __o2(g)? __co2(g) + __h2o(g)
forsale [732]
To balance this equation, first we should consider balancing C because it only presents in one reactant and one product.  Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2.  Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O.  So the total number of O in the products is 12+3 = 15.  O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O
3 0
2 years ago
Based on the following thermodynamic data, calculate the boiling point of ethanol in degrees Celsius. Substance ΔH∘f (kJ/mol) S∘
levacccp [35]

Answer:

76.03 °C.

Explanation:

Equation:

C2H5OH(l) --> C2H5OH(g)

ΔHvaporization = ΔH(products) - ΔH (reactants)

= (-235.1 kJ/mol) - (-277.7 kK/mol)

= 42.6 kJ/mol.

ΔSvaporization = ΔS(products) - ΔS(reactants)

= 282.6 J/K.mol - 160.6 J/K.mol

= 122 J/K.mol

= 0.122 kJ/K.mol

Using gibbs free energy equation,

ΔG = ΔH - TΔS

ΔG = 0,

T = ΔH/ΔS

T = 42.6/0.122

= 349.18 K.

Coverting Kelvin to °C,

= 349.18 - 273.15

= 76.03 °C.

8 0
2 years ago
The mechanism for electrophilic aromatic substitution reaction involves addition and
nika2105 [10]
I got it for you no worries

5 0
3 years ago
How are acids and ionic compounds similar?
11Alexandr11 [23.1K]

Answer: Ionic compounds are held together by the virtue of their opposing charges. Na+Cl- for example. If we consider Hg+(2Cl-)2, a mercuric chloride, the solubility is much less. Ba++(SO)4 Barium Sulphate, is highly insoluble; all differ by the relative attractiveness by Differing opposing charge(s).

Acids are very similar, consider Formic Acid, HCOOH, the simplest of the Carboxylic Acids. It dissociates more than say Benzoic Acid, C6H5-COOH. But neither disassociate as fully as Nitric Acid HNO3.

So the relative disassociation of the H+ (proton), or H3O+, (Hydronium ion), from any of these in water vary for a number of reasons we need not consider now.

Here is a “Tricky One!” (And very nasty). Take HF liquid or gas. This is one of the strongest acids on Earth - AS A LIQUID compound OR GAS. It will dissociate essentially near completion! Eat the floor, and is very dangerous.

NOW - HF (aqueous). The HF is in water. Very like HCl? NO! Why you may ask...The Electrophilic nature of Fluorine, “bathed in water, with an H+ all its own”, doesn’t let it go as easily!

HF is HIGHLY ordered in water, you can almost imagine a sort of “Hydrated matrix”, little HFs in endless rows...

BUT BE WARNED - even the aqueous HF is so reactive it will dissolve bone!

(I was told it was extremely painful; and did not appear to heal for weeks!)

Explanation: so, both types of compounds have a similarity, held together by the strength of their opposing charges or the degree of dissociation, (using water for simplicity).

That should do it.

8 0
2 years ago
Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)
Reptile [31]

<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

For the given chemical reaction:

C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0
3 years ago
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