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Nezavi [6.7K]
3 years ago
9

All of the following are true statements about catalysts except: Select one: a. A catalyst will speed the rate-determining step

b. A catalyst may induce steric strain in a molecule to make it react more readily c. A catalyst will lower the activation energy of a reaction d. A catalyst will be used up in the reaction
Chemistry
1 answer:
liberstina [14]3 years ago
6 0
<span>d. A catalyst will be used up in the reaction is wrong,
amount of catalyst should not change</span>
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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
What is the pOH of a 0.0072 M solution of sodium hydroxide? pOH=?
Rus_ich [418]
POH of a 0.0072 M=-lg(0.0072) = 2.1426675
8 0
3 years ago
Read 2 more answers
From the relative rates of effusion of ²³⁵UF₆ and ²³⁸UF₆ , find the number of steps needed to produce a sample of the enriched f
Dafna11 [192]

The number of steps required to manufacture a sample of the 3.0 mole%  ²³⁵U enriched fuel used in many nuclear reactors from the relative rates of effusion of ²³⁵UF₆ and ²³⁸UF₆. ²³⁵U occurs naturally in an abundance of 0.72% are :  mining, milling, conversion, enrichment, fuel fabrication and electricity generation.

<h3>What is Uranium abundance ? </h3>
  • The majority of the 500 commercial nuclear power reactors that are currently in operation or being built across the world need their fuel to be enriched in the U-235 isotope.
  • This enrichment is done commercially using centrifuges filled with gaseous uranium.
  • A laser-excitation-based method is being developed in Australia.
  • Uranium oxide needs to be changed into a fluoride before enrichment so that it can be treated as a gas at low temperature.
  • Uranium enrichment is a delicate technology from the perspective of non-proliferation and needs to be subject to strict international regulation. The capacity for world enrichment is vastly overbuilt.

The two isotopes of uranium that are most commonly found in nature are U-235 and U-238. The 'fission' or breaking of the U-235 atoms, which releases energy in the form of heat, is how nuclear reactors generate energy. The primary fissile isotope of uranium is U-235.

The U-235 isotope makes up 0.7% of naturally occurring uranium. The U-238 isotope, which has a small direct contribution to the fission process, makes up the majority of the remaining 99.3%. (though it does so indirectly by the formation of fissile isotopes of plutonium). A physical procedure called isotope separation is used to concentrate (or "enrich") one isotope in comparison to others. The majority of reactors are light water reactors (of the PWR and BWR kinds) and need their fuel to have uranium enriched by 0.7% to 3-5% U-235.

There is some interest in increasing the level of enrichment to around 7%, and even over 20% for particular special power reactor fuels, as high-assay LEU (HALEU).

Although uranium-235 and uranium-238 are chemically identical, they have different physical characteristics, most notably mass. The U-235 atom has an atomic mass of 235 units due to its 92 protons and 143 neutrons in its nucleus. The U-238 nucleus has 146 neutrons—three more than the U-235 nucleus—in addition to its 92 protons, giving it a mass of 238 units.

The isotopes may be separated due to the mass difference between U-235 and U-238, which also makes it possible to "enrich" or raise the proportion of U-235. This slight mass difference is used, directly or indirectly, in all current and historical enrichment procedures.

Some reactors employ naturally occurring uranium as its fuel, such as the British Magnox and Canadian Candu reactors. (By contrast, to manufacture at least 90% U-235, uranium needed for nuclear bombs would need to be enriched in facilities created just for that purpose.)

Uranium oxide from the mine is first transformed into uranium hexafluoride in a separate conversion plant because enrichment operations need the metal to be in a gaseous state at a low temperature.

To know more about Effusion please click here : brainly.com/question/22359712

#SPJ4

7 0
2 years ago
Maggie has 4 binders with 30 stamps in each binder. She has 3 binders with 36 baseball cards in each binder. If she gives away 4
Sliva [168]

Answer:

The number of stamps and cards Maggie has left if she gives 45 stamps to a friend is 183

Explanation:

If Maggie gives 45 stamps to a friend, you must calculate the number of stamps and cards she has left.

You know Maggie has 4 folders with 30 stamps in each folder. So the number of stamps she owns is calculated as:

4 folders*30 stamps in each folder= 120 stamps

If Maggie gives 45 stamps to a friend, then the number of stamps she has left will be calculated as the difference (the subtraction) between the stamps she owned and the ones she gives away:

120 stamps - 45 stamps= 75 stamps

On the other hand, she has 3 binders with 36 baseball cards in each binder. So the number of cards she owns is calculated as:

3 binders * 36 baseball cards in each binders= 108 baseball cards

The number of stamps and cards you have left is calculated as:

75 stamps + 108 baseball cards= 183

<em><u>The number of stamps and cards Maggie has left if she gives 45 stamps to a friend is 183</u></em>

4 0
2 years ago
β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa
gtnhenbr [62]

Answer:

The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=icRT

where,

\pi = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273.15 +25]=298.15 K

Putting values in above equation, we get:

0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L

The concentration of solute is 2.4278\times 10^{-5} mol/L

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

C=\frac{n}{V(L)}

n=2.4278\times 10^{-5} mol/L\times 0.137 L

n=3.3261\times 10^{-6} mol

To calculate the molecular mass of solute, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of β‑Galactosidase = 3.3261\times 10^{-6} mol

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

3 0
3 years ago
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