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Zinaida [17]
2 years ago
6

(Help would be greatly appreciated) What is the molarity of a solution made from 325.4g of AlCl3 with enough water to make 500.0

mL?
1. 4.88g
2. 4.88M
3. 2.440M
4. 2.44M
Chemistry
1 answer:
puteri [66]2 years ago
7 0

Answer:

4.88 M

Explanation:

M = number moles/Volume (L)

MM AlCl3 = 133.34 g/mol

Volume = 500 mL = 0.5 L

number moles = mass/MM = 325.4/133.34 = 2.44 moles AlCl3

M = 2.44/0.5 = 4.88 M

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A particular atom contains 29 electrons, 34 neutrons, and 29 protons. what is the identity of this element and what is its atomi
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Answer:

             Identity =  Copper

             Atomic Number =  29

Explanation:

Given Data:

Number of Electrons  =  29

Number of Neutrons =  34

Number of Protons  =  29

Identity and atomic number of Element:

                              This element can be identified by using two sources, i) Atomic Number ii) Electronic Configuration.

Atomic Number is defined as the number of protons present in the nucleus of an element. As the given number of protons are 29, therefore, the atomic number of this element is also 29.

Secondly, if the element is in neutral state then its number of protons must be equal to the number of electrons present in it. So, in neutral state this element will contain 29 electrons. Therefore, the electronic configuration is as follow,

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Now using electronic configuration one can easily identify the position of this element in periodic table. As the valence shell is 4, so it belongs to period 4 and there are 9 electrons in d-subshell so it belongs to transition elemnts and transition element with 9 electrons in d shell is Copper. Also, the stable electronic configuration of copper is written as,

                                     1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d¹⁰

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The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?
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Empirical formula is the simplest ratio of whole numbers of components in a compound. 
Assuming for 100 g of the compound 
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mass                      48.41 g                          19.02 g                      32.57 g
number of moles    48.41 / 63.5 g/mol      19.02 / 75 g/mol        32.57 / 32 g/mol 
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divide by the least number of moles 
                               0.762 / 0.2536             0.2536 / 0.2536         1.018 / 0.2536
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once they are rounded off 
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
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