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masya89 [10]
3 years ago
10

If a sodium atom lost one of its electrons and a chlorine atom gained it, what kind of bond would be formed between sodium and c

hlorine?
a. ionic
b. valence
c. metallic
d. covalent
Chemistry
2 answers:
Vikki [24]3 years ago
4 0
With no doubt, this type of bond would be ionic.

An ionic bond is when we transfer electrons. This means that the atoms would gain a charge.

Na+Cl- would be the resulting compound because:-

Na (sodium) loses : which means it would GAIN a positive charge.

Cl (chlorine) gains: which means it GAINS a negative charge.

Natasha_Volkova [10]3 years ago
3 0
This would be an ionic bond.  Ionic bonds form between metals and nonmetals due to the relatively large different in electronegativities where nonmetals are more electronegative than metals.  This difference in electronegativities makes it so that the metal gives up the electron to make an octet and the nonmetal gains the electron to make an octet (there is no sharing of the electron).  The metal becomes a cation while the nonmetal becomes a anion and their opposite electric charges keep them together.

I hope this helps let me know in the comments if anything is unclear.
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22.17 grams of Mercury (II) Nitrate, Hg(NO3)2, reacts with an excess of Potassium, K.
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Answer:

Mass = 14.0 g

Explanation:

Given data:

Mass of mercury nitrate = 22.17 g

Mass of mercury formed = ?

Solution:

Chemical equation:

Hg(NO₃)₂ + 2K    →    2KNO₃ + Hg

Number of moles of mercury nitrate:

Number of moles = mass/molar mass

Number of moles = 22.17 g / 324.6 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of Hg(NO₃)₂ and mercury.

                Hg(NO₃)₂       :        Hg

                     1                :          1

                  0.07             :       0.07

Mass of mercury:

Mass = number of moles × molar mass

Mass = 0.07 mol × 200.6 g/mol

Mass = 14.0 g

7 0
3 years ago
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For which compound does 0.26 mole weigh 13g
nalin [4]
0,26mol  ------- 13g
1mol       -------  x

x = (13g*1mol) / 0,26mol =  50g

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1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?
love history [14]

Answer:

m=1.325gNa_2CO_3

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

eq-g=0.5eq-g/L*0.050L=0.025eq-g

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3

Finally, the mass is computed by using its molar mass (106 g/mol)

m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3

Regards.

7 0
3 years ago
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