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1 year ago

Which of the titanium-containing minerals rutile (TiO2) or ilmenite (FeTiO3) has the larger percentage of titanium?

Chemistry

2 answers:

IRINA_888 [86]1 year ago

8 0

Both rutile (TiO2) and ilmenite (FeTiO3) has the equal percentage of titanium.

<u>Explanation</u>:

Titanium dioxide has the equal percentage of rutile and ilmenite. It's molecular weight is 79.87 gram per mole. Titanium oxide is a naturally occuring element produced from rutile,ilmenite and anatase.
These minerals combine and form titanium dioxide. It will be used in many applications. It will be used in many chemical reactions.
Titanium has a ph percentage of 7.5. It has a correct level of ph. So both has a equal level of titanium.

cupoosta [38]1 year ago

7 0

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.

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Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 10

vlada-n [284]

Answer:

a. both temperature changes will be the same

Explanation:

When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:

Q = m×C×ΔT

<em>Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.</em>

Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.

m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.

And Q is the heat released: If 2g release X heat, 4g release 2X.

Thus, ΔT in the experiments is:

Experiment 1:

X / 102C = ΔT

Experiment 2:

2X / 204C = ΔT

X / 102C = ΔT

That means,

<h3>a. both temperature changes will be the same</h3>

5 0

1 year ago

Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol

Contact [7]

1. The molarity of the stock solution of luminol = 1,431 M

2. 0.12 moles of luminol is present in 2.00 L of the diluted spray

3. The volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B): 83.86 ml

<h3>Further explanation</h3>

Stoichiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.

In stoichiometry included :

1. Relative atomic mass
2. Relative molecular mass

is the relative atomic mass of the molecule

3. mole

1 mole is the number of particles contained in a substance with the same number of atoms in 12 gr C-12

1 mole = 6.02.10²³ particles

While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element or the relative mass of the molecule

$\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}$

Luminol (C₈H₇N₃O₂) is a substance used to detect traces of blood in the scene of a crime because it reacts with iron in the blood

a. a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.

So the molarity is

1. Luminol mole

- the relative molecular mass of Luminol

= 8. C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams / mol

so the mole:

mol = gram / relative molecular mass

$mole=\frac{19}{177}$

mole = 0.1073

2. Molarity (M)

M = mole / volume

$M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}$

M = 1,431

b. luminol concentration in a spray bottle 6.00 × 10⁻² M. So that in 2 L of solution, the number of moles is:

mole = M x volume

mole = 6.10⁻² x 2

mole = 0.12

c. Molarity of the stock solution (Part A) = 1,431 M

the number of moles present in the diluted solution (Part B) = 0.12

So the volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B) is:

volume = mol / M

$volume\:=\:\frac{0.12}{1.431}$

volume = 0.08386 L = 83.86 ml

<h3>Learn more</h3>

moles of water you can produce

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the ratio of hydrogen atoms (H) to oxygen atoms (O) in 2 l of water

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Keywords: mole, volume, molarity, Luminol, the relative molecular mass

6 0

2 years ago

Read 2 more answers

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

1 year ago

Imagine you are given a mystery element. It is, however, a discovered and known element. You may perform a maximum of two observ

miv72 [106K]

The classification of it being a metal, nonmetal, or metalliod will be useful in the process of elimination to determine what it is. Then for the second test, meauring the atomin radius will narrow it down quicker to the mystery elemet's name.

Since you determined what part of the periodic table it's on, then when measuring the atomic radius, you should be able to pinpoint what the element is more surely.

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1 year ago

Read 2 more answers

Molar solubility of CuBr Ksp = 6.3 x 10-9

maksim [4K]

First, lets write the equation: CuBr--> Cu+2 + Br-
then, write the Ksp expression: Ksp= [Cu+2] [Br-]= 6.3 x 10-9
We let x be the concentration of Ksp that dissolved. Ksp= (x) (x)= 6.3 x 10-9.
Now we solve for x

X^2= 6.3x10-9
x= square root(6.3x10-9)
x= 7.94x10-5

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1 year ago