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3 years ago

Using a spring as an example, explain the compressions and rarefactions of a wave.

Chemistry

1 answer:

Vladimir [108]3 years ago

5 0

Answer:

Compressions are the parts where the coils are close together.

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How many atoms are in a casein molecule?

konstantin123 [22]

The heavy atom count of a casein molecule is 143 I believe

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3 years ago

Electron A falls from energy level X to energy level Y and releases blue light. Electron B falls from energy level Y to energy l

MrMuchimi

Answer: Transition from X to Y will have greater energy difference.

Explanation: For studying the energy difference, we require Planck's equation.

$E=\frac{hc}{\lambda}$

where, h = Planck's Constant

c = Speed of light

E = Energy

$\lambda$ = Wavelength of particle

From the equation, it is visible that the energy and wavelength follow inverse relation which means that with low wavelength value, energy will be the highest and vice-versa.

As electron A falls from X-energy level to Y-energy level, it releases blue light which has low wavelength value (around 470 nm) which means that it has high energy.

Similarly, Electron B releases red light when it falls from Y-energy level to Z-energy level, which has high wavelength value (around 700 nm), giving it a low energy value.

Energy Difference between X-energy level and Y-energy level will be more.

5 0

3 years ago

What are the parts of the atomic theory

Anvisha [2.4K]

Answer:

Explanation:

Dalton’s Atomic Theory

All elements are composed of tiny indivisible particles called atoms.

All atoms of the same element are identical.

Atoms of different elements are different(they have different sizes, masses, chemical properties, etc.).

Atoms of different elements can combine with each other in simple whole number ratios to form compounds. (Law of Definite Proportions).

Chemical reactions occur when atoms are separated, joined, or rearranged. However, atoms of one element are NOT changed into atoms of another element by a chemical reaction. (Law of Conservation of Mass).

4 0

3 years ago

KOH (aq) + Fe(NO3)3 (aq) →

atroni [7]

Answer:

see below

Explanation:

1. Predicting products (double replacement): ab + cd ---> ad + cb

KNO₃(aq) + Fe(OH)₃(s)

2. balance the equation

3KOH (aq) + Fe(NO3)₃ (aq) ---> 3KNO₃(aq) + Fe(OH)₃(s)

3. I don't know if you need this but ionic equation: only aqueous things get split into ions; gas, liquid, and solids stay together

3K⁺(aq) + 3(OH)⁻(aq) + Fe³⁺(aq) + 3NO₃⁻(aq) ---> 3K ⁺(aq) + 3NO₃⁻(aq) + Fe(OH)₃(s)

removing things on both product and reactant side

3(OH)⁻(aq) + Fe³⁺(aq) --->Fe(OH)₃(s)

7 0

3 years ago

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

3 years ago