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irinina [24]
3 years ago
7

Plz help me it a timed test

Chemistry
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

B

Explanation:

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Which of the following compounds contains the Lead (II) ion?<br> PbO<br> PbCl4<br> Pb2O<br> Pb2S
KATRIN_1 [288]
<span>PbO Let's look at each of the 4 compounds and see what's needed. PbO. * Oxygen has a valance shell that's missing 2 electrons and wants to get those 2 elections. Lead donates them, so you have a Lead (II) ions. This is a correct choice. PbCl4 * Chlorine wants to grab 1 electron to fill it's valance shell and Lead donates that election. However, there's 4 chlorine atoms and every one of them wants and electron, and lead is donating all 4 of the desired electrons making the Lead (IV) ion. So this is a bad choice. Pb2O * Oxygen still wants 2 electrons and gets them from the lead. But there's 2 lead atoms and each of them donates 1 election making for 2 Lead(I) ions. So this too is a bad choice. Pb2S * Sulfur is in the same column of the periodic table as oxygen and if this compound were to exist would have similar properties as Pb2O and would have Lead(I) ions. So this is a bad choice.</span>
6 0
3 years ago
Read 2 more answers
Regarding the Lyman series of hydrogen, calculate the frequency of the n = 5 line.
zvonat [6]
The Lyman series can be expressed in the formula <span><span>1/λ</span>=<span>RH</span><span>(1−<span>1/<span>n2</span></span>) where </span><span><span>RH</span>=1.0968×<span>107</span><span>m<span>−1</span></span>=<span><span>13.6eV</span><span>hc

</span></span></span></span>Where n is a natural number greater than or equal to 2 (i.e. n = 2,3,4,...). Therefore, the lines seen in the image above are the wavelengths corresponding to n=2 on the right, to n=∞on the left (there are infinitely many spectral lines, but they become very dense as they approach to n=∞<span> (Lyman limit), so only some of the first lines and the last one appear).
 The wavelengths (nm) in the Lyman series are all ultraviolet
:2 3 4 5 6 7 8 9 10 11
 Wavelength (nm) 121.6 102.6 97.3 95 93.8 93.1 92.6 92.3 92.1 91.9 91.18 (Lyman limit)
In your case for the n=5 line you have to replace "n" in the above formula for 5 and you should get a value of 95 x 10^-9 m for the wavelength. then you have to use the other equation that convert wavelength to frequency. </span>
4 0
3 years ago
What is the mass, in grams, of a 12.0cm³ sample of aluminum? The density of aluminum is 2.70g/cm³
Levart [38]

Answer:

The answer is

<h2>32.4 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of aluminum = 12 cm³

Density = 2.70 g/cm³

The mass of aluminum is

mass = 2.7 × 12

We have the final answer as

<h3>32.4 g</h3>

Hope this helps you

4 0
3 years ago
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astra-53 [7]

it's Lithium or Li

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5 0
3 years ago
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Jet001 [13]

Answer:

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7 0
3 years ago
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