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wolverine [178]
3 years ago
9

The whiye pigment TiO2 is prepared by the reaction of titanium tetrachloride, TiCl4, with water vapor in the gas phase:

Chemistry
1 answer:
omeli [17]3 years ago
5 0

Answer:

\boxed{\text{62.1 kJ}}

Explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})

                          TiCl₄(g) + 2H₂O(g) ⟶ TiO₂(s) + 4HCl(g)

ΔH°f/kJ·mol⁻¹:    -763.2     -241.828     -939.7    -92.307

\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [-939.7 + 4(-92.307)] - [-763.2 + 2(-241.828)\\& = & [-939.7 - 369.228] - [-763.2 - 483.656]\\& = & -1308.928 + 1246.856\\& = & \mathbf{-62.1}\\\end{array}\\\text{The amount of heat evolved is } \boxed{\textbf{62.1 kJ}}

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Paladinen [302]

Answer:

a boy pulling a toy train, the boy is

interacting with an object while applying a force to it,

another example of contact forces is friction.

Explanation:

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(btw love your pfp)

4 0
3 years ago
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The radioisotope radon-222 has a half life of 3.8 days. How much of a 10 gram sample of radon-222 would be left after 15.2 days?
djyliett [7]

Answer:

0.625 g

Explanation:

Given data:

Half life of radon-222 = 3.8 days

Total mass of sample = 10 g

Mass left after 15.2 days = ?

Solution:

Number of half lives = T elapsed / Half life

Number of half lives = 15.2 / 3.8

Number of half lives = 4

At time zero = 10 g

At first half life = 10 g/2 = 5 g

At 2nd half life = 5 g/ 2= 2.5 g

At third half life = 2.5 g/2 = 1.25 g

At 4th half life = 1.25 g/2 = 0.625 g

4 0
3 years ago
Suppose a liquid level from 5.5 to 8.6 m is linearly converted to pneumatic pressure from 3 to 15 psi. What pressure will result
wlad13 [49]

Answer:

a) P = 9.58 psi for  h=7.2 m

b) P=4.7 psi for h=5.94 m

Explanation:

Since the pressure Pon a static liquid level h is

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where p₀= initial pressure , ρ=density , g = gravity

then he variation of the liquid level Δh will produce a variation of pressure of

ΔP= ρ*g*Δh → ΔP/Δh =  ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m)  = 12/3.1 psi/m

if the liquid level is converted linearly

P = P₁ + ΔP/Δh*(h -h₁)

therefore choosing  P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m

P = 3 psi  + 12/3.1 psi/m *(7.2 m -5.5 m) = 9.58 psi

then P = 9.58 psi for  h=7.2 m

for P=4.7 psi

4.7 psi = 3 psi  + 12/3.1 psi/m *(h -5.5 m)

h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m

then P=4.7 psi for h=5.94 m

5 0
3 years ago
What should iron oxide be categorized
11111nata11111 [884]

Answer:

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Alenkasestr [34]

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