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riadik2000 [5.3K]
3 years ago
5

What is the study of the spectra produced from the emission and absorption of electromagnetic radiation? study of absorption spe

ctrums study of emission spectrum nanotechnology spectroscopy
Chemistry
2 answers:
Finger [1]3 years ago
5 0

Answer:

D spectroscopy

Explanation:

skelet666 [1.2K]3 years ago
3 0

Answer the correct answer out of the four is option c spectroscopy

Explanation- The interaction between electromagnetic radiation and matter studied. This study is commonly known as spectroscopy. It can also be named as study of absorption spectra or emission spectra.

The former spectrum is formed when energy is absorbed from Photons or light energy by electrons while the latter spectrum is formed due to a wavelength of light that is released when electrons jump from higher to lower level.  

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75km/hr

A car traveling with constant speed travels 150km in 7200 s. What is the speed of the car
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Which element has the highest ionization energy<br> A. Calcium<br> B. Bromine<br> C. Astatine
inessss [21]

Answer:

bromine

Explanation:

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An object 5 millimeters high is located 15 millimeters in front of a plane mirror. How far from the mirror is the image located?
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The correct answer for the question that is being presented above is this one: "D. 15 millimeters." An object 5 millimeters high is located 15 millimeters in front of a plane mirror. T<span>he image of the mirror is located 15 millimeters.</span>
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3 years ago
Gaseous ethane (CH,CH,) will react with gaseous oxygen (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0). Sup
BigorU [14]

<u>Answer:</u> The mass of CO_2 produced is 12.32 g

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For ethane:</u>

Given mass of ethane = 4.21 g

Molar mass of ethane = 30 g/mol

Putting values in equation 1, we get:

\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 31.9 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol

The chemical equation for the combustion of ethane follows:

2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O

By stoichiometry of the reaction:

If 2 moles of ethane reacts with 7 moles of oxygen gas  

So, 0.140 moles of ethane will react with = \frac{7}{2}\times 0.140=0.49mol of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, ethane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of ethane produces 4 moles of CO_2

So, 0.140 moles of ethane will produce = \frac{4}{2}\times 0.140=0.28mol of CO_2

We know, molar mass of CO_2 = 44 g/mol

Putting values in above equation, we get:

\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g

Hence, the mass of CO_2 produced is 12.32 g

7 0
2 years ago
Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose
Tom [10]

Answer:

Maximum amount of CO_{2} can be produced is 37.5 g

Explanation:

Balanced equation: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

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Molar mass of O_{2} = 32 g/mol

Molar mass of CO_{2} = 44.01 g/mol

So, 24 g of butane  = \frac{58.12}{24}mol of butane = 2.422 mol of butane

Also, 44.3 g of O_{2}  = \frac{44.3}{32}mol of O_{2} = 1.384 mol of O_{2}

According to balanced equation-

2 moles of butane produce 8 mol of CO_{2}

So, 2.422 moles of butane produce (\frac{8}{2}\times 2.422)moles of CO_{2} = 9.688 moles of CO_{2}

13 moles of O_{2} produce 8 mol of CO_{2}

So, 1.384 moles of O_{2} produce (\frac{8}{13}\times 1.384)moles of CO_{2} = 0.8517 moles of CO_{2}

As least number of moles of CO_{2} are produced from O_{2} therefore O_{2} is the limiting reagent.

So, maximum amount of CO_{2} can be produced = 0.8517 moles = (44.01\times 0.8517)g=37.5 g

4 0
3 years ago
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