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3 years ago

A catalyst:

Chemistry

2 answers:

svetlana [45]3 years ago

4 0

Answer:

Catalyst decreases activation energy

Explanation:

Consider the attached diagram and note the annotation => top of transition diagram for catalyzed reaction is lower than uncatalyzed reaction.

posledela3 years ago

3 0

Answer:

Explanation:

well, just need to remember

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PLEASE HELP!!!!!! WILL GIVE BRAINLIEST!!!! IT'S SUPER EASY

Nana76 [90]

the primary consumer in those photos would be C

6 0

3 years ago

The boiling point of chloroform is 61.7 C. The enthalapy of vaporization is 31.4 kj/mol. Caculate the entropy of vaporization. D

dedylja [7]

Answer:

Δ S = 93.8 J/mol-K

Explanation:

Given,

Boiling point of chloroform = 61.7 °C

= 273 + 61.7 = 334.7 K.

Enthalapy of vapourization = 31.4 kJ/mol.

Using Gibbs free energy equation

Δ G = Δ H - T (ΔS)

at equilibrium (when the liquid is boiling), Δ G = 0

so, 0 = ΔH - T (Δ S)

T (Δ S) = Δ H

and ΔS = ΔH / T

Δ S = (31400 J/mol.) / 334.7 K

Δ S = 93.8 J/mol-K

6 0

3 years ago

27. The density of nickel is 8.91 g/cm3. How large a cube, in cm3, would contain 2.00 x 10^24 atoms of nickel? Use dimensional a

jok3333 [9.3K]

Answer : The volume of the cube is, $21.88cm^3$

Solution : Given,

Density of nickel = $8.91g/cm^3$

Number of nickel atoms = $2\times 10^{24}$

Molar mass of nickel = 58.7 g/mole

First we have to calculate the moles of nickel.

As, $6.022\times 10^{23}$ atoms form 1 mole of nickel

So, $2\times 10^{24}$ atoms form $\frac{2\times 10^{24}}{6.022\times 10^{23}}=3.321$ moles of nickel

The moles of nickel = 3.321 moles

Now we have to calculate the mass of nickel.

$\text{ Mass of Ni}=\text{ Moles of Ni}\times \text{ Molar mass of Ni}$

$\text{ Mass of Ni}=(3.321moles)\times (58.7g/mole)=194.94g$

The mass of nickel = 194.94 g

Now we have to calculate the volume of nickel.

$Density=\frac{Mass}{Volume}$

$8.91g/cm^3=\frac{194.94g}{Volume}$

$Volume=21.88cm^3$

Therefore, the volume of the cube is, $21.88cm^3$

4 0

3 years ago

A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1.

Harrizon [31]

Answer:

27.60 g urea

Explanation:

The freezing-point depression is expressed by the formula:

ΔT= Kf * m

In this case,

ΔT = 5.6 - (-0.9) = 6.5 °C

Kf = 7.78 °C kg·mol⁻¹

m is the molality of the urea solution in X (mol urea/kg of X)

First we calculate the molality:

6.5 °C = 7.78 °C kg·mol⁻¹ * m

m = 0.84 m

Now we calculate the moles of urea that were dissolved:

550 g X ⇒ 550 / 1000 = 0.550 kg X

0.84 m = mol Urea / 0.550 kg X

mol Urea = 0.46 mol

Finally we calculate the mass of urea, using its molecular weight:

0.46 mol * 60.06 g/mol = 27.60 g urea

7 0

3 years ago

The formation of iodine is described by the following chemical equation:

olga nikolaevna [1]

Answer:

N2O2(g) +O2(g) ===> 2NO2(g)

Explanation:

For a nonelementary reaction, the reaction equation is described as the sum of all the steps involved. All these steps constitute the reaction mechanism. Each step in the mechanism is an elementary reaction. The rate law of the overall reaction involves the rate determining step (slowest step) in the reaction sequence.

Now look at the overall reaction 2NO(g) + O2(g) ---------> 2NO2(g)

The two steps in the mechanism are

2NO(g) --------->N2O2(g) (fast)

N2O2(g) +O2(g) -------> 2NO2(g) (slow)

Summing all the steps and cancelling out the intermediate N2O2(g), we obtain the reaction equation;

2NO(g) + O2(g) ---------> 2NO2(g)

Hence the answer.

7 0

3 years ago