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never [62]
2 years ago
14

The volume of an ideal gas is held constant. Determine the ratio P2/P1 of the final pressure to the initial pressure when the te

mperature of the gas rises (a) from 46 to 92 K and (b) from 35.4 to 69.0 oC.
Chemistry
1 answer:
cricket20 [7]2 years ago
5 0

Answer:

A. P₂ / P₁ = 2

B. P₂ / P₁ = 1.1

Explanation:

A. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 46 K

Final temperature (T₂) = 92 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/46 = P₂/92

Cross multiply

46 × P₂ = P₁ × 92

Divide both side by P₁

46 × P₂ / P₁ = 92

Divide both side by 46

P₂ / P₁ = 92 / 46

P₂ / P₁ = 2

B. Determination of the ratio P₂/P₁

Volume = constant

Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K

Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K

Final pressure /Initial pressure (P₂/P₁) =?

P₁/T₁ = P₂/T₂

P₁/308.4 = P₂/342

Cross multiply

308.4 × P₂ = P₁ × 342

Divide both side by P₁

308.4 × P₂ / P₁ = 342

Divide both side by 308.4

P₂ / P₁ = 342 / 308.4

P₂ / P₁ = 1.1

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Answer:

ΔH°r = -483.64 kJ

Explanation:

Let's consider the following balanced equation.

2 H₂(g) + O₂(g)  ⇒ 2 H₂O(g)

We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.

ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr

where

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n: moles

p: products

r: reactants

ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol

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ΔH°r = -483.64 kJ

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3 years ago
Sam believes the rock is more denser than the pencil. He measured the mass of the rock to be 8.5 grams and volume to be 4.5mL. H
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Answer:

Rock

Explanation:

Let's calculate the density of each object:

Rock:

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irina1246 [14]

Answer:

In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.

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How many calories is required to change the temperature of 2.18g of water from 15.3°C to 69.5°C. The specific heat of liquid wat
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The number  of calories that are  required  to change the temperature  of 2.18 g of water from 15.3 c to 69.5 c is  <u>118.16 cal</u>


    <u><em> calculation</em></u>

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