CHEMISTRY HIGH SCHOOL

Which two substances bind using a lock-and-key mechanism?enzyme and substrate
reactant and substrate
substrate and product
reactant and product

Answers

Answer 1
Answer: The answer is the first option. Enzyme and substrate bind using a lock-and-key mechanism. Enzymes act on a specific substrate and a substrate needs a specific enzyme, this is what is called a lock-and-key mechanism. Enzymes and substtates are like a key and a lock, one is for each other.
Answer 2
Answer: it would be A: enzyme and substrate

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How/why do copy machines produce o zone?

Answers

It's saying that it "protects us from the ozone layer." but too much may be harmful to your body because we aren't used to that 

Why can't there be a temperature lower than absolute zero?

Answers

At absolute zero, there would be no energy left, and negative energy is unknown to us, so you cannot go lower than a state with no energy. Atoms would stop moving.

A substance that can be separated into two or more substances only by a chemical change is

Answers

A substance that can be separated into two or more substances only by a chemical change is known as a heterogeneous mixture

A sample contains 10.5 g of the radioisotope Pb-212 and 157.5 g of its daughter isotope, Bi-212. How many half-lives have passed since the sample originally formed?4
14
15
147.5

Answers

Answer: The sample must have passed 4 half-lives after the sample was originally formed.

Explanation: This is a type of radioactive decay and all the radioactive process follow first order kinetics.

Equation for the reaction of decay of _(82)^(212)\textrm{Pb} radioisotope follows:

_(82)^(212)\textrm{Pb}\rightarrow _(83)^(212)\textrm{Bi}+_(-1)^0\beta

To calculate the initial amount of _(82)^(212)\textrm{Pb}, we will require the stoichiometry of the reaction and the moles of the reactant and product.


Expression for calculating the moles is given by:

\text{no of moles}=\frac{\text{Given mass}}{\text{Molar mass}}


Moles of _(82)^(212)\textrm{Pb} left = (10.5g)/(212g/mol)=0.0495moles  

Moles of _(83)^(212)\textrm{Bi}=(157.5g)/(212g/mol)=0.7429moles


By the stoichiometry of above reaction,


1 mole of _(83)^(212)\textrm{Bi} is produced by 1 mole _(82)^(212)\textrm{Pb}


So, 0.7429 moles of _(83)^(212)\textrm{Bi} will be produced by = (1)/(1)* 0.7429=0.7429\text{ moles of }_(82)^(212)\textrm{Pb}


Amount of _(82)^(212)\textrm{Pb}  decomposed will be = 0.7429 moles

Initial amount of _(82)^(212)\textrm{Pb}  will be = Amount decomposed + Amount left = (0.0495 + 0.7429)moles = 0.7924 moles

Now, to calculate the number of half lives, we use the formula:

a=(a_o)/(2^n)

where,

a = amount of reactant left after n-half lives = 0.0495 moles

a_o = Initial amount of the reactant = 0.7924 moles

n = number of half lives

Putting values in above equation, we get:

0.0495=(0.7924)/(2^n)

2^n=16.0080

Taking log on both sides, we get

n\log2=\log(16.0080)\nn=4

Answer:

4

Explanation:

Edg 2020

Consider the reaction C12H22O11(s)+12O2(g)→12CO2(g)+11H2O(l) in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/∘C. The temperature increase inside the calorimeter was found to be 22.0 ∘C. Calculate the change in internal energy, ΔE, for this reaction per mole of sucrose.

Answers

Answer:

the change in internal energy per mole of sucrose is 5989 KJ/mol

Explanation:

Since the calorimeter is a bomb calorimeter , where the volume remains constant

ΔE= Qrx

And assuming no heat losses to the surroundings

Qv + Qrx = 0

Cc*ΔT - Qrx = 0 ,

Qrx = Cc * ΔT = 7.50 KJ/°C * 22 °C = 165 KJ

the change in internal energy is calculated dividing by the number of sucrose moles involved n. the molecular weight of sucrose is

Mw= 12* 12g/mol + 22* 1g/mol + 11* 16 g/mol = 342 g/mol

n = m / Mw = 10 g / 342 g/mol = 0.029 mol

the change in internal energy per mole of sucrose is

Δe= ΔE/n = 165 KJ/0.029 mol = 5989 KJ/mol

What is true of the rate of radioactive decay? A. The rate is constant. B. The rate is continuously accelerating. C. The rate is continuously decelerating. D. The rate is exponentially accelerating.

Answers

The truest of the rate of radioactive decay is that the rate is exponentially accelerating. The answer is letter D. The rate is constant, the rate is continuously accelerating, the rate is continuously decelerating do not answer the question above.

Answer:

its a) the rate of decay is constant.

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