Which of the following is the smallest volume? 500 mL 2,500 cm3 5.5 x 10-1 L 25 m3

Why did Rutherford choose alpha particles in his experiment?

Georgia [21]

Rutherford used gold for his scattering experiment because gold is the most malleable metal and he wanted the thinnest layer as possible. The goldsheet used was around 1000 atoms thick. Therefore, Rutherford selected a Gold foil in his alpha scatttering experiment.

5 0

3 years ago

What is the energy of electromagnetic radiation with wavelength of 568 nm?

anygoal [31]

Apr 15, 2020 · 1 answer
How to get unblocked. 1. See answer. Add answer. Log in to add comment. lunlun02015 is waiting for your help. Add your

6 0

3 years ago

According to the collision theory and model created to explain the collision theory, what two factors must be satisfied for a gi

storchak [24]

The answer is correct orientation and sufficient energy. Answer choice D

4 0

3 years ago

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Distinguish between sugar and non-sugar with examples.

kap26 [50]

Answer:

Reducing sugars are sugars where the anomeric carbon has an OH group attached that can reduce other compounds. Non-reducing sugars do not have an OH group attached to the anomeric carbon so they cannot reduce other compounds. ... Maltose and lactose are reducing sugars, while sucrose is a non-reducing sugar

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

4 0

3 years ago