Rutile (TiO2) or ilmenite (FeTiO3) titanium-containing minerals has the larger percentage of titanium is explained below.
Explanation:
1. Titanium is obtained from different ores that occur naturally on the Earth. Ilmenite (FeTiO3) and rutile (TiO2) are the most important sources of titanium.
2. According to USGS, Ilmenite(FeTio3) accounts for about 92% of the world’s consumption of titanium minerals.
3. World resources of anatase, ilmenite and rutile total more than 2 billion tonnes. Identified reserves total 750 million tonnes (ilmenite plus rutile).
4. China, with 20 million tonnes--accounting for 29% of the world total-- is now the country that is most abundant in terms of ilmenite reserves. Meanwhile, Australia, with 24 million tones rutile reserves—accounting for 50% of the world total—is now the country that is most abundant in terms of rutile reserves.
5.Ilmenite can be mined from both layered intrusive deposits and heavy mineral deposits. It is often found alongside rutile in heavy mineral deposits. Ilmenite is used to make titanium dioxide pigment or it can be processed into feedstock that can be used in the manufacture of titanium. This has become more common as viable rutile deposits become increasingly scarce. South Africa and Australia are among the world’s largest producer of Ilmenite, each extracting over a million metric tonnes per year.
6.The Mining of titanium from intrusive rock is restricted to ilmenite and its weathered derivative leucoxene. The largest opencast ilmenite mine is Tellnes in Norway’s municipality of Sokndal.
Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
<span>Higher energy = shorter wavelength
Frequency is one cycle over an amount of time (seconds)
So higher frequency = higher energy = shorter wavelength</span>
