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3 years ago

A catalyst accelerates a reaction because:_____.

Chemistry

1 answer:

Margaret [11]3 years ago

3 0

Answer:

D. Lowering activation energy for the reaction

Explanation:

Catalysts increase the rate of a reaction without being used up. They do this by lowering the activation energy needed. With a catalyst, more collisions result in a reaction, so the rate of reaction increases.

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Match the term with its description. (4 points)

zlopas [31]

Answer:

2B, 1A, 3D, 4C, because of the definitions, they match.

4 0

3 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

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

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

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

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

Kito puts his spoon into a helping of fresh, hot, mashed potatoes as shown in the figure. Which process takes place when the two

Svet_ta [14]

Answer:

B. is the answer

Explanation:

Energy in form of heat is transferred from the warmer mashed potatoes to the cooler spoon.

7 0

3 years ago

A 312 g sample of a metal is heated to 277.845 °C and plunged into 200 g of water at a temperature of 11.945 °C. The final tempe

tigry1 [53]

Answer:

The specific heat capacity of the metal is 1.307 J/g °C

Explanation:

<u>Step 1:</u> Data given

mass of the metal = 312 grams

initial temperature of the metal ( before plunged in the water) = 277.845 °C

initial temperature water = 11.945 °C

Final temperature of water (and aslo the metal) = 99.062 °C

Specific heat capacity of wayer = 4.184 J/g °C

<u>Step 2:</u> Calculate the specifi heat capacity of the metal

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(metal) * C(metal) * ΔT(metal) = - m(water) * C(water) * ΔT(water)

with mass of metal = 312 grams

with C(metal) = TO BE DETERMINED

with ΔT (metal) = 99.062 - 277.845 = -178.783 °C

with mass of water = 200 grams

with C(water) = 4.184 J/°C * g

with ΔT (water) = 99.062 - 11.945 = 87.117 °C

312 * C(metal) * -178.783 = - 200* 4.184 * 87.117

C(metal) = -72899.51 / 55780.296 = 1.307

The specific heat capacity of the metal is 1.307 J/g °C

6 0

3 years ago

How was Aristotle’s model similar to Ptolemy’s model?

Vinvika [58]

The answer is that both models stated that Earth is the center of the system around which other objects orbit.

That is Aristotle’s model similar to Ptolemy’s model in the way that both stated that Earth is the center of the system around which other objects orbit.

Ptolemy’s model stated that Earth is the center of universe and other the sun, moon, planets, and stars revolving about it in circular orbits at increasing distance.

Aristotle’s model stated that the Sun, the Moon, the planets, and the stars travel in different separate spheres. And music of the spheres can be listen when the spheres touch each other. can be heard. He proved that the Earth is spherical,also stated that it is the the center of the universe.

6 0

3 years ago

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