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3 years ago

A student has an unknown sample. How can spectroscopy be used to identify the sample?

2 answers:

4 0

Every substance has a unique interaction or response to an electromagnetic radiation. He can use spectroscopy by looking at the response of the unknown sample to an electromagnetic radiation. Spectroscopy is is the study of the interaction between matter and electromagnetic radiation.

DiKsa [7]3 years ago

4 0

the answer is A!!!!!!!

You might be interested in

What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g

ArbitrLikvidat [17]

KClO3(s)+Δ→KCl(s)+32O2(g)

Approx. 3L of dioxygen gas will be evolved.

Explanation:

We assume that the reaction as written proceeds quantitatively.

Moles of KClO3(s) = 10.0⋅g122.55⋅g⋅mol−1 = 0.0816⋅mol

And thus 32×0.0816⋅mol dioxygen are produced, i.e. 0.122⋅mol.

At STP, an Ideal Gas occupies a volume of 22.4⋅L⋅mol−1.

And thus, volume of gas produced = 22.4⋅L⋅mol−1×0.0816⋅mol≅3L

Note that this reaction would not work well without catalysis, typically MnO2.

8 0

3 years ago

Read 2 more answers

If you had a sample containing 55 alpha particles, how many protons would be present in that sample? Show all necessary work.

arlik [135]

An alpha particle is a helium nucleus without the electrons - 2 neutrons and 2 protons.

So, all you would have to do is 2 x 55, which equals 110.

8 0

3 years ago

Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso

Sauron [17]

Answer:

For A: The number of arsenic atoms are $3.4\times 10^{21}$

For B: The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

Explanation:

For A:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of dimercaprol = 696 mg = 0.696 g (Conversion factor: 1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

$\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.0056 moles of dimercaprol will contain $0.0056\times 6.022\times 10^{23}=3.4\times 10^{21}$ number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, $3.4\times 10^{21}$ number of dimercaprol molecules will bind with = $1\times 3.4\times 10^{21}=3.4\times 10^{21}$ number of arsenic atoms

Hence, the number of arsenic atoms are $3.4\times 10^{21}$

For B:

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

$\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100$ ......(1)

For mercury:

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

$\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%$

For thallium:

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

$\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%$

For chromium:

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

$\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%$

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

8 0

3 years ago

42.9 liters of hydrogen are collected over water at 76.0 °C and has a pressure of 294 kPa. What would the pressure (in kPa) of t

BabaBlast [244]

Answer:

494.1 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 294 kPa

P2 = ?

V1 = 42.9 liters

V2 = 22.8 liters

T1 = 76.0°C = 76 + 273 = 349K

T2 = 38.7°C = 38.7 + 273 = 311.7K

294 × 42.9/349 = P2 × 22.8/311.7

12612.6/349 = 22.8 P2/311.7

36.14 = 22.8P2/311.7

Cross multiply

36.14 × 311.7 = 22.8P2

11264.605 = 22.8P2

P2 = 11264.605 ÷ 22.8

P2 = 494.1 kPa

3 0

2 years ago

A balloon is floating around outside your window. The temperature outside is 21 ∘C , and the air pressure is 0.700 atm . Your ne

Andreyy89

Answer:

$V=162L$

Explanation:

Hello,

In this case, we can compute the required volume by using the ideal gas equation as shown below:

$PV=nRT$

Thus, solving for the volume and considering absolute temperature (in Kelvins), we obtain:

$V=\frac{nRT}{P}= \frac{4.70mol*0.082\frac{atm*L}{mol*K}*(21+273)K}{0.700atm} \\\\V=162L$

Best regards.

4 0

3 years ago