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IrinaK [193]
2 years ago
9

Identify each element below, and give the symbols of the other elements in its group:

Chemistry
1 answer:
Misha Larkins [42]2 years ago
4 0

Answer:

Answer in explanation

Explanation:

a. Boron , element 5

Helium has 2 electrons, add to the other 3 to give 5.

Other group members are : Aluminum Al, Gallium Ga, Indium In , Thallium Tl and Nihonium Nh

b. Sulphur, element 16

Neon is 10 , add other 6 electrons to make 16

Other group members are: Oxygen O, selenium Se , Tellurium Te and Polonium Po

c. Lanthanum, element 57

Xenon is 54, add the other 3 electrons to give 57.

Other elements in group : Scandium Sc , Yttrium Y , Actinium Ac, Lutetium Lu and/or Lawrencium Lr

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Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed leave it blank. If no reaction occurs leav
babunello [35]

Answer:

H+  +  OH−  --> H2O

Explanation:

Hydrochloric acid is represented by the chemical formular; HCl. This is an ionic substance so in water it breaks apart into hydrohrn ions; H+ and chloride ions; Cl−. It is a strong acid, hence it completely dissociates.

Potassium Hydroxide is also an ionic substance it also breaks apart in water into potassium ions; K+ and hydroxide ions; OH−. It is a strong base, hence it completely dissociates.

The complete ionic equation for the reaction is given as;

H+  +  Cl−  +  K+  +  OH−  -->  K+  +  Cl−  +  H2O

The Hydrogen ion and the Hydroxide ions combine to form water.

The net ionic equation is given as;

H+  +  OH−  --> H2O

Cl- and K+ ions were cancelled out because they do not undergo any changes therefore are not part of the net ionic equation. They are referred to as spectator ions.

5 0
2 years ago
A sample of barium nitrate is placed into a jar containing water. The mass of the barium nitrate sample is 27g. Assume the water
34kurt
So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.

using that relation
i.e.
8.7g (barium nitrate) =100g (water)
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27g barium nitrate = (100/ 8.7 ) × 27
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5 0
3 years ago
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3 years ago
A solution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 235 mL of solution. How many moles o
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Answer:

The correct answer is 0.206 moles

Explanation:

According to the given scenario, the calculation of the number of moles of ammonium chloride is available in the resulting solution is given below:

Given that

Amount of NH_4Cl is 11.0 grams

And, the volume is 235 mL

Now the molar mass of NH_4Cl is 53.49g/mol

So, the number of moles presented is

= 11.0 ÷ 53.49

= 0.206 moles

hence, the number of moles of ammonium chloride are available in the resulting solution is 0.206 moles

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2 years ago
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Answer:

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