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Alchen [17]
3 years ago
7

Es un cambio natural o provocado, donde se produce energía para agua y dióxido de carbono: a)Precipitación. b)Fermentación c) Ef

ervecencia. d) Combustión.
Chemistry
1 answer:
Andre45 [30]3 years ago
6 0

Answer:

d) Combustión.

Explanation:

¡Hola!

En este caso, dado que estamos enfocados en el concepto de cambio químico, el cual se caracteriza por exhibir un cambio en la composición e identidad de las sustancias iniciales (reactivos) a otras finales (productos).

Ahora bien, como se nos dice que los productos de este cambio químico son energía, agua y dióxido de carbono, inferimos que el nombre de este proceso es d) Combustión, por ejemplo la combustion del gas natural para calentar nuestras comidas en la cocina.

¡Saludos!

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Some water released 7800 J of energy when cooled from 78°C to 33°C. What was the mass of the water?
Karo-lina-s [1.5K]

Answer:

-41. 47

Explanation:

m = q / Cp x T

m = Mass

q = Energy (or joules)

Cp = Heat Capacity

T = Change in Temperature

Water's heat capacity is always 4.18.

This is the formula you'll need for change in temperature:

Final - Initial

So, 33 - 78 = -45

m = 7800 / 4.18 x -45

= -41.47

4 0
2 years ago
For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
Alik [6]

Answer : The rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-2}M^{-2}s^{-1}

Thus, the value of the rate constant 'k' for this reaction is 7.5\times 10^{-2}M^{-2}s^{-1}

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1

\text{Rate}=4.25\times 10^{-2}Ms^{-1}

Therefore, the rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

3 0
3 years ago
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Alborosie

Answer:

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Explanation:

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Answer:

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Explanation:

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H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, \Delta H_{hydro} is higher (less negative) for trans-1,3-pentadiene

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4 years ago
calculate the volume of carbon dioxide of room temperature and pressure obtained from 30 grams of glucose​
ivanzaharov [21]

Answer:

A

Explanation:

4 0
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