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Juli2301 [7.4K]
4 years ago
15

How many atoms of titanium are there in 0.125 mole of each of the following?a. ilmenite, FeTiO3b. titanium(IV) chloridec. Ti2O3d

. Ti3O5
Chemistry
1 answer:
Gekata [30.6K]4 years ago
7 0

Answer:

a. 7.528x10²² atoms Ti

b. 7.528x10²² atoms Ti

c. 1.5055x10²³ atoms Ti

d. 2.258x10²³ atoms Ti

Explanation:

Using the molecular formula of the structures, we can determine the moles of titanium in 0.125 moles of each compound. With moles of titanium we can convert these to atoms using Avogadro's number as follows:

a. FeTiO₃: 1 mole of Ti per mole of FeTiO₃.

Moles of Ti are 0.125 moles. Atoms are:

0.125 moles Ti * (6.022x10²³ atoms / 1 mole) =

<h3>7.528x10²² atoms Ti</h3>

b. Titanium (IV) chloride = TiCl₄: 1 mole of Ti per mole of TiCl₄.

Moles of Ti are 0.125 moles. Atoms are:

0.125 moles Ti * (6.022x10²³ atoms / 1 mole) =

<h3>7.528x10²² atoms Ti</h3>

c. Ti₂O₃: 2 moles of Ti per mole of Ti₂O₃.

Moles of Ti are 2*0.125 moles = 0.25 moles. Atoms are:

0.25 moles Ti * (6.022x10²³ atoms / 1 mole) =

<h3>1.5055x10²³ atoms Ti</h3>

a. Ti₃O₅: 3 mole of Ti per mole of Ti₃O₅.

Moles of Ti are 3*0.125 moles = 0.375 moles of Ti. Atoms are:

0.375 moles Ti * (6.022x10²³ atoms / 1 mole) =

<h3>2.258x10²³ atoms Ti</h3>
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Consider the following system at equilibrium:
Vika [28.1K]

Answer:

1) Rightward shift

2) Rightward shift

3) Leftward shift

4) Leftward shift

5) Leftward shift

6) Rightward shift

7) No shift

8) No shift                                                              

   

Explanation:

To evaluate each case we need to consider Le Chatelier's Principle, which states that the adding of additional reactants or products to a system will shift the equilibrium in the opposite direction, to maintain the equilibrium of the system. On the contrary, if we remove a reactant or a product in the system, the equilibrium will be shifted in the direction of the reactant or product reduced, to produce more of it (and thus maintain balance).        

Taking into account the above, let's see each statement, in the following equation:

A(aq) + B(aq)  ⇄  2C(aq)    (1)

1) Increase A. This will cause a rightward shift in equation 1 in order to consume the reactant added.

2) Increase B. Same as 1), this will cause a rightward in equation 1.

3) Increase C. This will cause a leftward shift in order to consume the excess of product in the system.  

4) Decrease A. This will produce a leftward shift to produce the reactant that is being reduced.

5) Decrease B. Same as 4), a leftward shift.

6) Decrease C. This will produce a rightward shift to produce the product that is being reduced.

7) Double A, half B. The double A will cause a rightward shift and the half B will produce a leftward shift, which results in no shift.

8) Double both B and C. Double B will produce a rightward shift and double C will produce the contrary, a leftward shift, so the final result is no shift.

               

I hope it helps you!

4 0
4 years ago
A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
In-s [12.5K]

Answer:

3.6667

Explanation:

<u>For helium gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

{P_1}=1.8667\ atm

<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

{P_1}=1.8\ atm

<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

6 0
3 years ago
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