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2 years ago

D. Explain why hydrogen produces three light in spectrometer

Chemistry

1 answer:

horrorfan [7]2 years ago

8 0

Answer:

Though a hydrogen atom has only one electron, it contains a large number of shells, so when this single electron jumps from one shell to another, a photon is emitted, and the energy difference of the shells causes different wavelengths to be released. Hence, mono-electronic hydrogen has many spectral lines.

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Ms. Pitt set up a classroom demonstration to show what happens when a sample of zinc metal is add a.ed to a flask containing hyd

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volume of HCl remains the same

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2 years ago

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A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is th

Kay [80]

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

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mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

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3 years ago

schepotkina [342]

Answer:

b. 6.02 x 1023 molecules

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The number of moles in 27.6g ammonia is 27.6 ÷ 17 = 1.62 mol.

A mole is 6.02 × 10²³, so the number of hydrogen atoms in a 1.62 moles of ammonia is 1.62 × 6.02 × 10²³ × 3 = 2.93 × 10² atoms.

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2 years ago

Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen

alexgriva [62]

Answer:

$V_{base}=24.04mL$

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Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

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We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

$2M_{base}V_{base}=M_{acid}V_{acid}$

Therefore, for is to compute the volume of the used base, we proceed as shown below:

$V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}$

And we plug in to obtain:

$V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL$

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3 years ago

Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

sergejj [24]

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$ .

Since, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of $Al(C_{2}H_{3}O_{2})_{3}$ .

Thus, according to molecular formula, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.

1 atom of oxygen will be present in $\frac{1}{3.61\times 10^{24}}$ moles of $Al(C_{2}H_{3}O_{2})_{3}$ . Thus,

$2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}$ .

Molar mass of $Al(C_{2}H_{3}O_{2})_{3}$ is 236 g/mol, mass can be calculated as follows:

$m=n\times M=0.7285 mol\times 236 g/mol=172 g$

Therefore, mass of $Al(C_{2}H_{3}O_{2})_{3}$ will be 172 g.

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3 years ago

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D. Explain why hydrogen produces three light in spectrometer​

D. Explain why hydrogen produces three light in spectrometer