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Korolek [52]
3 years ago
15

If an 900.0 g sample of radium-226 decays to 225.0 g of radium-226 remaining in 3,200 years, what is the half-life of radium-226

? (3 points)

Chemistry
1 answer:
Stels [109]3 years ago
3 0

Answer:

1600 yr

Explanation:

The half-life of radium-226 is the time it takes for half of it to decay.  

After one half-life, half of the original amount will remain.  

After a second half-life, half of that amount will remain, and so on.  

We can construct a table as follows:  

\begin{array}{cccc}\textbf{No. of} &\textbf{Fraction} &\textbf{Mass}\\ \textbf{Half-lives} & \textbf{Remaining}&\textbf{Remaining/g}\\0 & 1 &900.0\\\\1 & \dfrac{1}{2} &450.0\\\\2 & \dfrac{1}{4} & 225.0\\\\3 & \dfrac{1}{8} & 112.5\\\\\end{array}

We see that the mass will drop to 225.0 g after two half-lives.

The mass dropped to 225.0 g in 3200 yr.  

If 3200 yr = 2 half lives,

1 half-life = 1600 yr

The decay curve for your sample is shown below. The mass has dropped to half its original value (450 g) after 1600 yr and to one -fourth  (225.0 g) after 3200 yr.

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Serjik [45]

The solubility equilibrium of CaCrO_{4}:

[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
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Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.

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A community consists of all the organisms in an ecosystem that belong to the same species.
choli [55]

Answer:

False

Explanation:

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What is the total energy change for the following reaction:CO+H2O-CO2+H2
Alekssandra [29.7K]

Answer:

\large \boxed{\text{-41.2 kJ/mol}}

Explanation:

Balanced equation:    CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)

(a) Enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}

(b) Total enthalpies of reactants and products

\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}

(c) Enthalpy of reaction \Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}

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bekas [8.4K]

the chemical equation will be XY2

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What is the molarity of a solution made by adding 0.720 mole of NaOH to 2.40 liters of water
WITCHER [35]

Answer:

Molarity = 0.3 M

Explanation:

Given data:

Moles of NaOH = 0.720 mol

Volume of water = 2.40 L

Molarity = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Molarity = 0.720 mol / 2.40 L

Molarity = 0.3 mol/L

Molarity = 0.3 M

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