Is potassium borate soluble?

After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?

Bond [772]

Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2) $^{t/t(1/2)}$ where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days

5 0

2 years ago

What is C6H5 called?

zysi [14]

C6H5 is the molecular formula for Phenyl.

5 0

2 years ago

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What is the molarity of a 10 L solution containing 5.0 moles of solute?

Setler79 [48]

Molarity = Moles of solute/ L(liters) of solution

So let's plug in the information.

5.0 moles/10L = 0.5 M

3 0

2 years ago

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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 $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago

Suppose a helium-3 nuclide transforms into a helium-4 nuclide by absorbing a proton and emitting a positron. Complete the nuclea

saw5 [17]

<u>Answer:</u> The nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.

<u>Explanation:</u>

Nuclear reaction are defined as the reactions in which nucleus of an atom is involved.

Positron emission is defined as the emission process in which positron particle is emitted. In this process, a proton gets converted to neutron and an electron neutrino particle.

$_Z^A\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e$

The chemical equation for the reaction of He-3 with a proton follows:

$_2^3\textrm{He}+_1^1\textrm{H}\rightarrow _2^4\textrm{He}+_{+1}^0e$

Hence, the nuclear equation for the conversion of He-3 nuclide to He-4 nuclide is given above.

6 0

2 years ago