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notsponge [240]
2 years ago
9

The molar mass of copper(II) chloride (CuCl2) is 134.45 g/mol. How many formula units of CuCl2 are present in 17.6 g of CuCl2? 7

.88 × 1022 formula units 1.84 × 1023 formula units 1.91 × 1023 formula units 1.42 × 1024 formula units
Chemistry
2 answers:
diamong [38]2 years ago
8 0

The answer is,

<u>A. 7.88 × 1022 formula units</u>



Sedaia [141]2 years ago
5 0
7.88  10^22 formula units 
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The model below shows an atom of an element. 10 light gray and 8 dark gray balls sit at the center with 2 concentric black rings
MA_775_DIABLO [31]

Answer:

The Atomic Number of the atom of an element whose model is given is "8" that is option no. 'C' in the question.

Explanation:

An Atom comprises of 3 basic structures that are Protons, Neutrons and Electrons. The central part is the Nucleus which contains protons and neutrons having positive charge and no charge respectively. The electrons are revolving around the nucleus in electronic shells having the negative charge.

<u><em>ATOMIC NUMBER: </em></u>

Atomic number is the number of protons present inside the nucleus of an atom and it determines the place of that particular atom in the <u>Periodic Table.</u>

In the model, given in the question, the nucleus contains 2 types of balls dark gray colored and light gray colored. The key at the bottom shows the dark gray colored ball as having a positive charge and thus it represents the atomic number for the given atom of element which is <u><em>EIGHT (8).</em></u>

So, the atomic number for the given atom is 8 which is element OXYGEN.  

9 0
2 years ago
Read 2 more answers
Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?
Virty [35]

Thee question is incomplete; the complete question is;

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Select one or more:

A: Chlorine

B: Carbon

C: Hydrogen

D: Sulfur

E: Fluorine

F: Oxygen

Answer:

chlorine

sulphur

Explanation:

The octet rule states that, for  atoms to be stable, they must have eight electrons on their outermost shells.

This rule is not strictly followed  by some elements such as sulphur and chlorine. The atoms of these elements can sometimes expand their octet by utilizing the d-orbitals found in the third principal energy level and beyond.

These leads to formation of compounds in which the central atom has more than eight electrons in its outermost shell.

5 0
2 years ago
A crystal of graphite contains which type of bond?
cupoosta [38]

Answer:Graphite has a giant covalent structure in which: each carbon atom is joined to three other carbon atoms by covalent bonds.

4 0
1 year ago
Read 2 more answers
Determine the total number of moles of ions in 40.0 ml of a 0.345 m solution of the strong electrolyte mgcl2
Sedbober [7]
A strong electrolyte like MgCl2 dissociates completely as per the following reaction:

MgCl_2 ----\ \textgreater \   Mg^{2+} + 2Cl^-

As you can see, from 1 molecule of MgCl2 produces 3 ions on dissociation.
So, 1 mole of MgCl2 produces 3 moles of ions.

Now, Moles of MgCl2 = Volume x Molarity

= 0.04 x 0.345   [Change volume to Litres]
= 0.0138 moles

Now, total moles of ions =  0.0138 x 3 = 0.0414
6 0
2 years ago
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
2 years ago
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