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liq [111]
3 years ago
8

The reaction of iron (III) metal with a solution of copper (II) sulfate releases iron ions into the solution through a single di

splacement reaction.
a. Determine the moles of iron ions produced in this reaction.

b. Name a soluble compound that could be added to precipitate all of the iron ions from the solution.

c. What mass of the soluble compound from part (c) is required to precipitate all of the iron ions you determined in part (b)? HINTs: determine a new chemical reaction with your soluble compound Fe2(SO4)3. Then, use the moles of Fe2(SO4)3 calculated in part (b)
Chemistry
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

. Name a soluble compound that could be added to precipitate all of the iron ions from the solution.

Sodium Hydroxide.

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The answer will be 2. 10.g
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At 250°C an equilibrium mixture of SbCl3(g), Cl2(g), and SbCl5(g) has the partial pressures 0.670 bar, 0.438 bar, and 0.228 bar,
Sveta_85 [38]

Explanation:

As the given reaction will be SbCl_{3}(g) + Cl_{2}(g) \rightarrow SbCl_{5}

Since, it is given that volume of reaction vessel is doubled. Hence, moles of species involved will also increase.

Therefore, the reaction equation will become as follows.

             2SbCl_{3}(g) + 2Cl_{2}(g) \rightarrow 2SbCl_{5}

As it is given that partial pressure of SbCl_{3} is 0.670 bar, Cl_{2} is 0.438 bar and SbCl_{5} is 0.228 bar.

Expression to calculate new equilibrium pressure if the volume of reaction vessel is double is as follows.

                  K_{p} = \frac{[P_{SbCl_{5}}]^{2}}{[P_{SbCl_{3}}]^{2}[P_{Cl_{2}}]^{2}}

                                   = \frac{(0.228)^{2}}{(0.670)^{2}(0.438)^{2}}

                                   = 6.05 \times 10^{3}

Thus, we can conclude that new equilibrium pressures if the volume of the reaction vessel is doubled is 6.05 \times 10^{3}.

4 0
3 years ago
What is the net amount of heat this person could radiate per second into a room at 18.0 ∘C (about 64.4 ∘F) if his skin's surface
marysya [2.9K]

Answer:

142.6 joules is the amount of heat that a person could radiate per second

Explanation:

To solve this, we have to apply Stephan-Boltzmann's law:

Q/ΔT = σ . ε . A . (Te⁴ - Ta⁴) where

σ = Boltzmann's constant → 5.67×10⁻⁸ W/m² . K⁴

ε = Body's emissivity, in this case = 1

Α = Surface, we assume a value of 2m²

Te = Temperature of the body's surface, in this case 30°C

Ta = Temperature of the room, where the body is. In this case, 18°C

Notice that T° must be Absolute T° → T°C + 273

18°C + 273 = 291K

30°C + 273 = 303K . Let's replace data:

Q/s = 5.67×10⁻⁸ W/m² . K⁴ . 1 . 2m² (303⁴K - 291⁴K)

Q/s = 5.67×10⁻⁸ W/m² . K⁴ . 1 . 2m² . 1.26×10⁻⁹K⁴

Q/s = 142.6 W

1 W  = Joules/s

4 0
3 years ago
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