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3 years ago

Please help me balance:_NaNO3 + _PbO_ ➡️ Pb(NO3)2 + _Na2O

Chemistry

1 answer:

KengaRu [80]3 years ago

7 0

Answer:

The balanced reaction is given by,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

Explanation:

The reaction is as given.

Lets count the number of each elements in the reaction.

In reactant side, number of sodium atoms are 1 , lead are 1, nitrogen are 1 and oxygen are 4.

in product side, number of sodium atoms are 2 , lead are 1 , nitrogen are 2 and oxygen are 7.

So we need to balance sodium and oxygen atoms in the reaction.

There is deficient of sodium and oxygen atoms on reactant side.

Thus, multiply (NaNO3) by 2.

Thus, sodium atoms become 2 , nitrogen 2 and oxygen 6. Total 7 oxygen atoms.

Thus, the balanced reaction is,

$2NaN(O)3 + PbO$ ⇒ $Pb(NO3)2 + Na2O$

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Given the equation representing a system at equilibrium N2(g)

RoseWind [281]

Answer:

Decreasing the temperature of the system will shift the reaction rightward.

Explanation:

The complete question is:

Given the equation representing a system at equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + energy

what changes occur when the temperature of this system is decreased?

<h2>Solution</h2>

Modifying the temperature of a system in equilibrium changes the equilibrium constant and the equilibrium position (concentrations) of the system.

When the temperature is decreased, following LeChatelier's principle that the system will react in a way that seeks to counteract the disturbance, the reaction will shift toward the reaction that produces more heat energy to compensate the temperature decrease.

Thus, decreasing the temperature of the system will favor the forward reaction, more N₂(g) and H₂(g) will be consumed and more NH₃(g) and energy will be produced. Hence, the equilibrium will shift rightward.

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3 years ago

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wave

timofeeve [1]

Explanation:

It is given that,

The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,

$n_i=8$

$\lambda=3745\ nm$

The amount of energy change during the transition is given by :

$\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

And

$\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]$

Plugging all the values we get :

$\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5$

So, the final level of the electron is 5.

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3 years ago

How to calculate the number of neutrons in the boron atom

Gekata [30.6K]

6 neutrons

Hope this helps :)

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3 years ago

Read 2 more answers

Calculate your percent yield for the bromination of cis-stilbene. [Note: Assume that stilbene is the limiting reagent. You must

evablogger [386]

Answer:

See explanation below

Explanation:

To calculate any percent yield of a reaction you need to provide, the theorical data and the experimental data, which you are not providing.

I will do an example with some values I found on another place to explain to you how to do it. You then, replace the data you have and follow the same procedure.

Let's suppose we have the following data of the stilbene:

density: 1.0111 g/mL

volume used: 0.3 mL

Molecular weight: 180.25 g/mol

Now, we use 0.3 mL of cis stilbene to do a reaction with acid and bromine to produce the 1,2-dibromo-1,2-diphenylethane.

The problem stated that the cis stilbene is the limiting reactant, therefore, the moles consumed of stilbene, would be the moles produced of the final product.

With the density let's calculate the mass of stilbene, and then, with the molecular weight, the moles:

d = m/V ---> m = d*V

m = 1.011 * 0.3 = 0.3033 g

moles = 0.3033 / 180.25 = 0.0017 moles

These obtained moles would be the moles of the final product too, because stilbene is the limiting reactant so:

moles of product = 0.0017 moles

Let's calculate the mass:

Molecular weight of 1,2-dibromo-1,2-diphenylethane = 339.8 g/mol

m = 0.0017 * 339.8 = 0.5776 g

This would be the theorical mass obtained in the experiment. Now, let's suppose we obtained a mass of 0.4158 g. This is the actual yield of the reaction, so the percetn yield would be:

%yield = Exp yield / theo yield * 100

Replacing:

%yield = 0.4158/0.5776 * 100

%yield = 71.99 %

This would be the %yield of the bromination. All you have to do now, is replace your theorical and experimental data and you should get to the final and accurate yield.

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