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3 years ago

Help please

Chemistry

1 answer:

Mekhanik [1.2K]3 years ago

7 0

Answer:

it's the regolith

hope this helps you

C Horizon - Also called regolith: the layer beneath the B Horizon and above the R Horizon. It consists of slightly broken-up bedrock. Plant roots do not penetrate into this layer; very little organic material is found in this layer. R Horizon - The unweathered rock (bedrock) layer that is beneath all the other layers.

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ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU + w

q = 475 J + 93 J

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Therefore, the magnitude of q for the process 568 J.

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Read 2 more answers

Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:

12345 [234]

Answer:

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

Explanation:

Given reaction: 2S + 3O₂ → 2 SO₃

Given: The enthalpy of reaction: ΔH = - 792 kJ

Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol

In the given reaction, the number of moles of S reacting: n = 2

As, Number of moles: $n = \frac{mass\: (w_{1})}{molar\: mass\: (m)}$

∴ mass of S in 2 moles of S: $w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g$

Given reaction: 2S + 3O₂ → 2 SO₃

In this reaction, the limiting reagent is S

⇒ 2 moles S produces (- 792 kJ) heat.

or, 64 g of S produces (- 792 kJ) heat.

∴ 42.8 g of S produces (x) amount of heat

⇒ The amount of heat produced by 42.8 g S:

$x = \frac{(- 792\: kJ) \times 42.8\: g}{64\: g} = (-529.65)\: kJ$

$\Rightarrow x = (-5.2965 \times 10^{2})\: kJ = (-5.2965 \times 10^{5})\: J$

$(\because 1 kJ = 10^{3} J)$

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

8 0

3 years ago