List some good ideas for Science Fair for Sixth Graders

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

cupoosta [38]

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.

7 0

3 years ago

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stomach acid contains dilute hydrochloric acid write a chemical equation to show the neutralization of stomach acid by milk of m

Eddi Din [679]

Answer:

Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq)

Explanation:

4 0

3 years ago

s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen g

Elenna [48]

This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

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

$\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol$

The given chemical reaction is:

$2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)$

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce = $\frac{3}{2}\times 0.05=0.0750mol$ of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of hydrogen gas = $21^oC=[21+273]K=294K$

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

$743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L$

Hence, the volume of hydrogen gas that will be collected is 1.85 L

8 0

3 years ago

How many Oxygen atoms in: 5H2SO3

astra-53 [7]

Answer:

20

Explanation: I think

5 0

2 years ago

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Determine the mass of sodium carbonate required to produce 23.4g of sodium chloride when it reacts with excess hydrochloric acid

Leni [432]

The balanced chemical reaction for the described reaction above is,
Na2CO3 + 2HCl ---> 2NaCl + H2CO3
From the reaction, 1 mole of Na2CO3 is needed to produce 2 moles of NaCl. In terms of mass, 106 grams of Na2CO3 are needed to produce 116.9 grams of NaCl. From this,
(23.4 g NaCl) x (106 g Na2CO3 / 116.9 NaCl = 21.22 g Na2CO3
Thus, approximately 21.22 g Na2CO3 is needed for the desired reaction.

8 0

3 years ago