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24 days ago

You have a balloon filled with hydrogen gas which keeps it at a

Chemistry

1 answer:

lorasvet [2.5K]24 days ago

5 0

The resulting temperature is 46.5°C.

Details:

According to Charles's law, the volume of gas, while maintaining constant pressure, correlates directly with temperature in Kelvin.

The formula representing Charles's law is expressed as follows:

$$\frac{V}{T} = constant$

$$\frac{V1}{T1} = \frac{V2}{T2}$

We need to determine T2, thus:

$$T2 = \frac{V2T1}{V1}$

V1 = 736 ml = 0.736 L

T1 = 15 ° C

V2 = 2.28 L

Substituting the values gives us:

T2 = $$\frac{2.28 \times 15}{0.736}$

= 46.5°C

It is evident that as the volume increases, the temperature also rises.

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Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O

alisha [2704]

Answer:

1219.5 kJ/mol

Explanation:

The calculation for this value requires using the following equation:

ΔHºrxn = Σn * (BE of reactants) - Σn * (BE of products)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H)].

The bond energy (BE) values are:

BE C = C: 839 kJ/mol

BE C-H: 413 kJ/mol

BE O = O: 495 kJ/mol

BE C = O: 799 kJ/mol

BE O-H: 463 kJ/mol

By substituting these values into the equation, you will get:

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1 month ago

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Alekssandra [2711]

The rate equation for a first order reaction can be expressed as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

In this context, k represents the reaction's rate constant, t denotes the time the reaction takes, $A_{0}$ is the initial concentration, and $A_{t}$ is the concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) If we start with an initial concentration of 100, when 90% of the substance is eliminated, the remaining quantity at time t will be 100-90=10. By substituting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

The time required to destroy 90% of the substance amounts to 23.03 days.

(b) If the initial concentration is set at 100, when 99% is destroyed, the present amount at time t will be 100-99=1. By substituting the input values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

This results in a duration of 46.06 days required to eradicate 99% of the chemical.

(c) Should the initial concentration be set at 100, with 99.9% of the chemical removed, the remaining quantity at time t will be 100-99.9=0.1. Substituting the values yields

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, the time needed to eliminate 99.9% of the chemical is calculated as 69.09 days.

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12 days ago