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Shkiper50 [21]
4 years ago
12

Carbon has two naturally occurring isotopes: C-12 (natural abundance is 98.93%) and C-13 (natural abundance is 1.07%). How many

C-13 atoms are present, on average, in a 30000-atom sample of carbon?
Chemistry
1 answer:
Sav [38]4 years ago
5 0
The number of atoms present, on average, will be the natural abundance of the isotope times the number of atoms in the sample => number of C-13 atoms = C-13 abundance * number of atoms in the sample = 1.07% * 30,000 = 1.07 * 30,000 / 100 = 321 atoms.<span> Answer: 321 atoms.</span>
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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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3 years ago
The mole fraction of nitrogen in the air is 0.7808. this means that 78.08% of the molecules in the air are nitrogen. when the at
Pachacha [2.7K]

Answer:

p = \boxed{\text{593 torr}}

Explanation:

For this question, we must use Dalton's Law of Partial Pressures:

The partial pressure of a gas in a mixture of gases equals its mole fraction times the total pressure:

p = \chi p_{\text{tot}}

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χ = 0.7808

p_{\text{tot}} = \text{ 760 torr}

Calculation:

p = 0.7808 \times \text{ 760 torr}\\\\p= \boxed{\textbf{593 torr}}

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