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Brrunno [24]
3 years ago
7

Calculate the wavelength of yellow light produced by a sodium lamp if the frequency of radiation is 3.34 x 10^14 Hz

Chemistry
1 answer:
max2010maxim [7]3 years ago
3 0

Answer :  The wavelength of yellow light produced by a sodium lamp is, 8.98\times 10^{-7}m

Explanation : Given,

Frequency of radiation = 3.34\times 10^{14}Hz=3.34\times 10^{14}s^{-1}

conversion used : Hz=s^{-1}

Formula used :

\nu=\frac{c}{\lambda}

where,

\nu = frequency of radiation

\lambda = wavelength of radiation

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

3.34\times 10^{14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}

\lambda=8.98\times 10^{-7}m

Therefore, the wavelength of yellow light produced by a sodium lamp is, 8.98\times 10^{-7}m

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Answer: -105 kJ

Explanation:-

The balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]

\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]

\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]

Delta H=-105kJ

Therefore, the enthalpy change for this reaction is, -105 kJ

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3 years ago
What happens to the glucose molecule during the process of cellular respiration? (5 points)
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The correct answer is option a, that is, it gets broken down.  

A set of metabolic reactions and procedures, which occurs in the cells of organisms to transform biochemical energy from nutrients into ATP, and then discharge waste components is known as cellular respiration. At the time of cellular respiration, a molecule of glucose gets dissociated slowly into water and carbon dioxide. With it, some of the ATP is generated directly in the reactions, which transform glucose.  


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Determine the acid dissociation constant for a 0.0250 M weak acid solution that has a pH of 2.37 . The equilibrium equation of i
Oksanka [162]

Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

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2 years ago
How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The t
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A solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water. The molar mass of Cu is 63.55 g/mol the molar mass of S
slamgirl [31]

The molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.

<h3>How to calculate molarity?</h3>

The molarity of a solution can be calculated using the following formula:

Molarity = no of moles/volume

According to this question, a solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water.

no.of moles of CuSO4 = 35g ÷ 159.6g/mol

no. of moles of CuSO4 = 0.22 moles

Therefore; molarity of CuSO4 solution is calculated as follows:

M = 0.22 ÷ 0.25

M = 0.88M

Therefore, the molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.

Learn more about molarity at: brainly.com/question/12127540

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