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Black_prince [1.1K]
3 years ago
11

How do you think you results would be affected if using K2Cr2O7/H2SO4 as the oxidizing agent instead of KMnO4

Chemistry
1 answer:
makkiz [27]3 years ago
3 0

Answer:

Due to weak oxidizing agent.

Explanation:

In my opinion the results would be affected if using K2Cr2O7/H2SO4 as the oxidizing agent instead of KMnO4 because K2Cr2O7/H2SO4 is weak oxidizing agent as compared to KMnO4. An oxidizing agent is a substance that has the ability to oxidize other substances means that accept their electrons so that's why the results of strong oxidizing agent is different than weak oxidizing agent.

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Using PV = nRT, we can calculate the moles of the sample.
874 mmHg = 116,524 Pa
n = PV/RT
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moles = mass / Mr
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Mr = 27.2
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Repeat units = 27.2 / 14 ≈ 2

Formula of substance:
C₂H₄

Combustion equation:
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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]

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

Explanation:

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Data  

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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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