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ch4aika [34]
3 years ago
10

A laser pointer emits light that has a wavelength of 5.32x 10–7 m. What is the frequency of the light?

Chemistry
1 answer:
butalik [34]3 years ago
8 0

Answer:

Frequency = 0.6 × 10¹⁵ s⁻¹

Explanation:

Given data:

Wavelength = 5.32 × 10⁻⁷ m

Frequency = ?

Solution:

Formula:

speed of light = wavelength × frequency

Frequency = speed of light / wavelength

Frequency =  3 × 10⁸ m/s /  5.32 × 10⁻⁷ m

Frequency = 0.6 × 10¹⁵ s⁻¹

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What mass of carbon dioxide is produced from the complete combustion of 5.30x10-3 g of methane?
pshichka [43]

Answer:

1.45 x 10⁻² g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

5.30 x 10⁻³ g CH₄          1 mole              1 mole CO₂          44.007 g
---------------------------  x  ----------------  x  ---------------------  x  -----------------  =
                                      16.043 g            1 mole CH₄            1 mole

=  0.0145 g CO₂

=  1.45 x 10⁻² g CO₂

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7 0
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What is chemical reaction?<br>​
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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
3 years ago
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