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5 months ago

If a 1.0x10-5 L sample of agas at 2.0x106 atm isreleased until it is equal to0.275 atm, what would thenew volume of the gas be?

Chemistry

1 answer:

Greeley [361]5 months ago

5 0

Answer:

Explanation:

Here, we want to get the new volume of the gas

What we need to know is the law that connects volume and pressure at constant temperature

The law that supports this is the Boyle's law

It states that the volume of a given mass of gas is inversely proportional to its pressure at constant temperature

Mathematically, we can have this as represented as:

$\begin{gathered} P_1V_1=P_2V_2 \\ P_1\text{ = 2.0 }\times10^{6\text{ }}\text{ atm} \\ V_1\text{ = 1.0 }\times10^{-5}\text{ L} \\ P_2\text{ =0.275 atm} \\ V_2\text{ = ?} \end{gathered}$

We can proceed mathematically to solve as follows;

$undefined$

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At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank

hodyreva [135]

Answer:

$\large \boxed{\text{-92 $^{\circ}$C}}$

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

$\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}$

2. Temperature in degrees Celsius

$\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$

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A reduction reaction involves either the addition of hydrogen or removal of:

Marina CMI [18]

Reduction involves the either the addition of hydrogen and removal of oxygen.

<h3>What is reduction?</h3>

Reduction involves the removal of oxygen.

This implies there is a loss of oxygen in reduction.

This can be represented in the extraction of iron from it ores.

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Reduction is also the addition of hydrogen. This implies it is the gain of hydrogen.

For example

CH₃CHO → CH₃CH₂OH

learn more on reduction here: brainly.com/question/9485345

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1 year 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

Klio2033 [76]

Rate equation for first order reaction is as follows:

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

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

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

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

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

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

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

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

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

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

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