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Anna [14]
3 years ago
15

How much calcium can be recovered from 75.0g of calcium fluoride

Chemistry
1 answer:
7nadin3 [17]3 years ago
3 0

38.46g

Explanation:

Given parameters:

Mass of CaF₂ = 75.0g

Unknown:

Mass of calcium that can be recovered = ?

Solution:

This is a mass percentage problem and we need to solve it accordingly.

To solve this problem;

Find the molar mass of  CaF₂

Find the ratio between the molar mass of Ca and that of  CaF₂

Multiply by the given mass

Molar mass of  CaF₂ = 40 + (2 x 19) = 78g/mol

Mass of calcium = \frac{40}{78} x 75 = 38.46g

learn more:

Mass percentage brainly.com/question/8170905

#learnwithBrainly

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Hydrogen ion, strictly, the nucleus of a hydrogen atom separated from its accompanying electron. The hydrogen nucleus is made up of a particle carrying a unit positive electric charge, called a proton. The isolated hydrogen ion, represented by the symbol H+, is therefore customarily used to represent a proton.
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Diamond is an allotrope of :
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carbon

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What is the molarity of a KF(aq) solution containing 3.0 mol of KF in 2.00L of solution?
sergeinik [125]

Answer : The molarity of KF in the solution is 1.5 M.

Explanation : Given,

Moles of KF = 3.0 mol

Volume of solution = 2.00 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }KF}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.0mol}{2.00L}=1.5mole/L=1.5M

Therefore, the molarity of KF in the solution is 1.5 M.

4 0
3 years ago
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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A substance that is impenetrable by x-rays is described as being radiopaque.

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Instead, they absorb or block the rays and when used in radiology, they appear white or light gray on photographic films.

Radiopaque materials are applied in generating ultrasound images and other forms of clinical procedures.

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