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djverab [1.8K]
2 years ago
13

5F248+ 2NH3(8) ► NF418 + 6HF

Chemistry
1 answer:
mylen [45]2 years ago
6 0

Answer:

14.2L at STP

Explanation:

Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:

<em>Moles NH3 -Molar mass: 17.031g/mol-</em>

3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3

<em>Moles HF:</em>

0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF

<em>Volume HF</em>

PV = nRT; V = nRT/P

<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>

Replacing:

V = 0.634moles*0.082atmL/molK*273.15K / 1atm

V = 14.2L at STP

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4 0
3 years ago
When 20.0 g of KI are dissolved in 50.0 mL of distilled water in a calorimeter, the temperature drops from 24.0 °C to 19.0 °C. C
Reil [10]
<h2>Answer:</h2>

<em>8.67kJ/mol</em>

<h2>Explanations</h2>

The formula for calculating the amount of heat absorbed by the water is given as:

\begin{gathered} q=mc\triangle t \\ q=50\times4.18\frac{J}{g^oC}\times(19-24) \\ q=50\times4.18\times(-5) \\ q=-1045Joules \\ q=-1.045kJ \end{gathered}

Determine the moles of KI

\begin{gathered} moles\text{ of KI}=\frac{mass\text{ of KI}}{molar\text{ mass of KI}} \\ moles\text{ of KI}=\frac{20g}{166g\text{/mol}} \\ moles\text{ of KI}=0.1205moles \end{gathered}

Since heat is lost, hence the enthalpy change of the solution will be negative that is:

\begin{gathered} \triangle H=-q \\ \triangle H=-(-1.045kJ) \\ \triangle H=1.045kJ \end{gathered}

Determine the enthalpy of solution in kJ•mol-1

\begin{gathered} \triangle H_{diss}=\frac{1.045kJ}{0.1205mole} \\ \triangle H_{diss}\approx8.67kJmol^{-1} \end{gathered}

Hence the enthalpy of solution in kJ•mol-1 for KI is 8.67kJ/mol

4 0
1 year ago
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