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Natalka [10]
3 years ago
7

Using the equations and enthalpy values provided, which mathematical expression can be used to determine the unknown enthalpy ch

ange represented by x?
3C + 3O2 yields 3CO2 deltaH = -1182 kJ

4H2 + 2O2 yields 4H2O deltaH =-1144 kJ

C3H8 + 5O2 yields 3CO2 + 4H2O deltaH = x

3H8 3C + 4H2 H = 104 kJ

A. 104 = (-1182) - [x + (-1144)]

B.104 = x - [(-1182) + (-1144)]

C.104 = (-1182) + (-1144) + x

D.104 = x + (-1182) - (-1144)
Chemistry
2 answers:
natita [175]3 years ago
5 0

Answer:

B. 104 = x - [(-1182) + (-1144)]

Explanation:

The unknown entahlpy change for the given equation is determined by the Hess's Law.

3C + 3O2 → 3CO2  ΔH = -1182 kJ      ....(a)

4H2 + 2O2  → 4H2O  ΔH =-1144 kJ   ....(b)

C3H8 → 3C + 4H2  ΔH  = 104 kJ  ....(c)

C3H8 + 5O2  →  3CO2 + 4H2O   ΔH = x   ....(d)

The unknown enthalpy change x is calculated as follows,

The equation (d) is obtained by the adding equation (a), equation (b ) and equation (c) .Therefore, the value of the unknown enthalpy change x will be equal to the sum of all enthalpy change value of the respective equation.

104 = x - [(-1182) + (-1144)]

vlabodo [156]3 years ago
3 0
In order to solve this, we need to make use of Hess' Law.

We are already given the equations and their corresponding deltaH. Using Hess' Law, we can generate this equation:
104 kJ = x - (-1182 kJ) - (-1144 kJ)

Among the choices, the answer is
<span>B.104 = x - [(-1182) + (-1144)] 
</span>
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The ionic equations represent the reactions between four metals, P, Q, R and S, and solutions of the salts of the same metals
EleoNora [17]

Answer:

P + Q2+ -> no reaction

R + P2+ -> R2+ + P

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S + Q2+ -> no reaction

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5 0
3 years ago
Enough of a monoprotic acid is dissolved in water to produce a 1.64 M solution. The pH of the resulting solution is 2.82 . Calcu
Lady bird [3.3K]

Answer:

Ka = 1.39x10⁻⁶

Explanation:

A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

<em>Where Ka is:</em>

Ka = [H⁺] [X⁻] / [HX]

<em>Where [] is the molar concentration in equilibrium of each specie. </em>

The equilibrium is reached when some HX reacts producing H+ and X-, that is:

[HX] = 1.64M - X

[H⁺] = X

[X⁻] = X

As pH is 2.82 = -log [H⁺]:

[H⁺] = 1.51x10⁻³M:

[HX] = 1.64M - 1.51x10⁻³M = 1.638M

[H⁺] = 1.51x10⁻³M

[X⁻] = 1.51x10⁻³M

And Ka is:

Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]

<h3>Ka = 1.39x10⁻⁶</h3>
6 0
3 years ago
PLEASE HELP!!!
neonofarm [45]

Answer:

Cl_2+2NaI\rightarrow I_2+2NaCl

Explanation:

Hello there!

In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:

Cl_2+NaI\rightarrow I_2+NaCl

It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:

Cl_2+2NaI\rightarrow I_2+2NaCl

Best regards!

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