You are asked to make 500 mL of a 0. 250 M sodium chloride (NaCl) solution.

Chemistry

1 answer:

nignag [31]2 years ago

3 0

Answer:

a. .125 mol b. 7.25g

Explanation:

a. To get the moles multiply the volume by the molarity.

.5L*.25M=.125mol NaCl

b. The molar mass is provided so multiply the moles by the molar mass to get the amount of grams of nacl.

.125mol*58molar mass=7.25g nacl to make the solution

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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.

4 0

3 years ago

10. A 2.36-gram sample of NaHCO3 was completely decomposed in an

Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

$m_r=m_{NaHCO_3}=2.36 g$