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Fed [463]
3 years ago
7

Magnesium has 2 valence electrons, and oxygen has 6 valence electrons. Which type of bonding is likely to occur between a magnes

ium atom and an oxygen atom?
Chemistry
1 answer:
Greeley [361]3 years ago
3 0
<span>In bonding with another element, magnesium has an oxidation number of +2. It gives away 2 electrons when it bonds. It will most easily bond with an element that accepts 2 electrons. [ With an oxidation number of -2 that is oxygen, which contains 6 electrons in its valence shell. </span>
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4.00 moles of CU(CN)2<br><br>Find the number of grams ​
UNO [17]

Hey there!

Cu(CN)₂

Find the molar mass.

Cu: 1 x 63.546 = 63.546

C: 2 x 12.01 = 24.02

N: 2 x 14.07 = 28.14

-----------------------------------

                      115.706 grams

The mass of one mole of Cu(CN)₂ is 115.706 grams.

We have 4 moles.

115.706 x 4 = 463

4.00 moles of Cu(CN)₂ has a mass of 463 grams.

Hope this helps!

6 0
3 years ago
if you dilute 25.4mL of a 3.5M solution to make 166.7mL of solution, what is the molarity of the dilute solution
Anna11 [10]
Molarity=(initial molarityxinitial volume)/final volume
M=(3.5x25.4)/166.7=0.533 M
8 0
3 years ago
Sugar can be removed from a sugar water solution through dialysis.
Fantom [35]

Answer:

false

which I think

I am not sure

4 0
3 years ago
In a particular titration experiment a 30.0 ml sample of an unknown hcl solution required 25.0 ml of 0.200 m naoh for the end po
Nikitich [7]
The balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
the number of NaOH moles reacted - 0.200 mol/L x 0.0250 L = 0.005 mol
according to molar ratio
number of NaOH moles reacted = number of HCl moles reacted 
therefore number of HCl moles - 0.005 mol 
volume of 30.0  mL contains 0.005 mol
therefore 1000 mL contains - 0.005 mol / 0.030 L = 0.167 M
concentration of HCl is 0.167 M
5 0
3 years ago
A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the
Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid  

HA + NaOH ==> NaA + H2O where HA is benzoic acid  

At the equivalence point, all the benzoic acid ==> sodium benzoate  

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)  

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11  

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150  

x^2 = 3.33x10^-12  

x = 1.8x10^-6 = [OH-]  

pOH = -log [OH-] = 5.74  

pH = 14 - pOH = 8.26  


5 0
3 years ago
Read 2 more answers
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