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3 years ago

For #11 how to determine the signs?

Chemistry

2 answers:

vladimir1956 [14]3 years ago

6 0

They a all positive bro

Viefleur [7K]3 years ago

3 0

The first reaction is exothermic as it releases heat as shown by the '+ 48 kJ'. When a reaction is exothermic the enthalpy change (ΔH) is negative as the system is losing heat. The entropy change (ΔS) is also negative as 4 moles of gas are reacting to form 2 moles of gas, reducing the disorder in the system. Since the spontaneity of a reaction is governed by ΔG = ΔH - TΔS, where ΔG must be negative for a spontaneous reaction, the reaction is only spontaneous at low temperatures as both ΔH and ΔS are negative, so as temperature increases, so does ΔG, toward 0.
The second reaction is endothermic as it absorbs heat therefore having a positive ΔH, and a positive ΔS as one mole of liquid is converted to one mole of more disordered gas. Using the same equation as before it can be shown that the reaction is only spontaneous at higher temperatures. These answers all hold true for the 3rd reaction as well (i.e. positive ΔH, positive ΔS and spontaneous at high temperatures).
Finally, reaction 4 is exothermic, therefore negative ΔH, and has a positive ΔS as 2 moles in solution are produced from one mole of solid. Therefore using the equation shown before, ΔG will always be negative regardless of temperature, and therefore the reaction is always spontaneous.
Hope this helps! If you could I would greatly appreciate you making this the brainliest answer :) If you have any questions feel free to ask.

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What is the heat in joules required to melt 25 grams of ice?

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Answer:

8,350 Joules

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2 years ago

What is molecular formula of Ozone?

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3 years ago

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Compounds A, B, and C react according to the following equation. 3A(g) + 2B(g) 2C(g) At 100°C a mixture of these gases at equili

Vedmedyk [2.9K]

Answer:

The value of Kc for the reaction is 3.24

Explanation:

A reversible chemical reaction, indicated by a double arrow, occurs in both directions: reagents transforming into products ( direct reaction) and products transforming back into reagents (inverse reaction)

Chemical Equilibrium is the state in which direct and indirect reactions have the same reaction rate. Then taking into account the rate constant of a direct reaction and its inverse the chemical constant Kc is defined.

Being:

aA + bB ⇔ cC + dD

where a, b, c and d are the stoichiometric coefficients, the equilibrium constant with the following equation:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reagents also raised to their stoichiometric coefficients.

Then, in the reaction 3A(g) + 2B(g) ⇔ 2C(g), the constant Kc is:

$Kc=\frac{[C]^{2} }{[A]^{3} *[B]^{2} }$

where:

[A]= 0.855 M
[B]= 1.23 M
[C]= 1.75 M

Replacing:

$Kc=\frac{1.75^{2} }{0.855^{3}*1.23^{2} }$

Solving you get:

Kc=3.24

<u><em>The value of Kc for the reaction is 3.24</em></u>

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3 years ago

What word or two-word phrase best describes the shape of the hydrogen cyanide ( hcn ) molecule?

elena-14-01-66 [18.8K]

Answer : Linear

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The molecule has two electron pairs in all and appears to be linear.

Also, according to the VSEPR theory; the electron clouds on atoms around the carbon will try to repel each other.

They will get pushed apart, which gives HCN molecule a linear molecular geometry or shape.

The bond angle that is developed will be 180 degrees since it has a linear molecular geometry of HCN. The hybridisation observed in this molecule is SP.

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3 years ago

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If u (x) = negative 2 x squared and v (x) = StartFraction 1 Over x EndFraction, what is the range of (u circle ma poottay

KonstantinChe [14]

The range is negative numbers.
The interval for the range is .
***You might want to look at your functions again because I don't see a choice that matches.
Step-by-step explanation:
Given functions:

We are asked to find the range of .
I'm also going to look at the domain just to see if this possibly might change my range .
is the inner function. So we will consider the domain of that function first.
You only have to worry about division by zero for the function .
Since we are dividing by , we don't want to be zero.
So far the domain is all real numbers except .
Now let's move out.
exists for all numbers, . So we didn't want to include from before.
Now let's put it together:

So the domain is still all real numbers except at since we cannot divide by 0 and is 0 when .
with .
is positive for all numbers except .
So is negative for all numbers since negative divided by positive is negative.
So the range is only negative numbers.
Let's also look at the inverse:

Multiply both sides by :

Divide both sides by :

Take the square root of both sides:
.
So can't be 0 and it also can't be positive because the inside of the square root will be negative (since negative divided by positive results in negative).

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3 years ago