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2 years ago

How do the terms denature, catalyst, active site, and substrate relate to each other

Chemistry

1 answer:

postnew [5]2 years ago

5 0

Explanation:

The enzyme 's active site binds to the substrate. Increasing the temperature generally increases the rate of a reaction, but dramatic changes in temperature and pH can denature an enzyme, thereby abolishing its action as a catalyst. ... When an enzyme binds its substrate it forms an enzyme-substrate complex.

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What Is the answers?

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Answer:

oxygen is responsible for rusting

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2 years ago

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Balance each of these equations.

Oksanka [162]

Answer:

A. 2NO + O2 -> 2NO2

B. 4Co + 3O2 -> 2Co2O3

C. 2Al + 3Cl2 -> Al2Cl6

D. 2C2H6 + 7O2 -> 4CO2 + 6H2O

E. TiCl4 + 4Na -> Ti + 4NaCl

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2 years ago

Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/

Ivenika [448]

Answer:

The answer to your question is: C₃H₃O This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

44 g -------------------- 12 g

3.2007 g ------------ x

x = (3.2007 x 12) / 44

x = 0.8729 g of Carbon

12 g of C -------------- 1 mol

0.8729 g -------------- x

x = (0.8729 x 1) / 12

x = 0.0727 mol of Carbon

For Hydrogen

18 g ---------------------- 1 g

1.3102 g ------------------- x

x = (1.3102 x 1) / 18

x = 0.0727 g of Hydrogen

1 g ------------------------ 1 mol

0.0727g ---------------- x

x = (0.0727 x 1)/1

x = 0.0727 mol of Hydrogen

For oxygen

g of Oxygen = g of sample - g of Carbon - g of hydrogen

g of Oxygen = 1.3109 - 0.8709 - 0.0727

g of Oxygen = 0.3673

16 g of Oxygen ------------- 1 mol of O

0.3673 g --------------------- x

x = (0.3673 x 1)/ 16

x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon 0.0727 / 0.023 = 3.1 ≈ 3

Hydrogen 0.0727 / 0.023 = 3.1 ≈ 3

Oxygen 0.0230 / 0.023 = 1

C₃H₃O

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3 years ago

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A rigid vessel contains water at 0.15 MPa (at 25 °C). The vessel is heated to the critical point of water. Calculate the fractio

pishuonlain [190]

Answer:

0.8078 Kg

Explanation:

Pressure of water = 0.15 MPa = 1.5 bar

At critical point of water ,temperature = 647 K=374°C

From the ideal gas equation

P×V= m×R×T

Let us assume volume = 1 m^3

1.5 x 105 x 1 = m x 287 x 647

m= 0.8078 kg

the fraction of mass of liquid at 25°C.

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3 years ago

A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo

Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>

Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

= (2.684 - 2.156) g

= 0.528 g

Find the number of moles of the elements in the compound

The number of moles is given by

$n = \frac{m}{M}$

where m = given mass and

M = Molar mass

Number of moles of Zinc = $\frac{2.156}{65}$ = 0.033 moles

Number of moles of Oxygen = $\frac{0.528}{16}$ = 0.033 moles

Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

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1 year ago