<span>The lithosphere is mostly made of Earth's outer layer, the crust, and the upper portion of Earth's mantle. The asthenosphere is the upper part of Earth's mantle "which is also the middle layer of Earth".

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6 0

3 years ago

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The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T th

Contact [7]

Answer:

-99.8 kJ

Explanation:

We are given the methodology to answer this question, which is basically Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.

The standard heat of reaction is

ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their standard states.

Compound ΔfHº (kJmol⁻¹)

SO₂ -296.8

O₂ 0

SO₃ -395.8

The balanced chemical equation is

SO₂(g) + ½O₂(g) → SO₃(g)

Thus

Δr, 298K Hº( kJmol⁻¹ ) = 1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹

Now the heat capacity of reaction will be be given in a similar fashion:

Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants

where ν is as above the stoichiometric coefficient in the balanced chemical equation.

Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90

= -3.90 JK⁻¹mol⁻¹

Finally Δr,500 K Hº = Δr, 298K Hº + CprxnΔT

Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8

= -99,787.8 J x 1 kJ/1000 J = -99.8 kJ

Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.

3 0

3 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

Balance the equation below. _NH4OH + __H3PO4 =>_(NH4)3PO4 + _H2O​

Balance the equation below. _NH4OH + __H3PO4 =>_(NH4)3PO4 + _H2O