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Bas_tet [7]
4 years ago
8

I need help with number 3. is the answer 1,2,3,or4

Chemistry
1 answer:
lubasha [3.4K]4 years ago
8 0

Answer:

1 and 4

Explanation:

in both pictures the temperature is cooling down

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A) a column

example: earth alkaline metals
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46. True or false. Chemical reactions do not involve changes in the chemical bonds that join
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Answer:

Chemical reactions do not involve changes in the chemical bonds that join

atoms in compounds :

<u>False</u>

Explanation:

Chemical reaction are the reaction in which old bonds break and new  bonds are formed . The formation of new bond result in formation of new compounds . This happen because new bond are result of linking different atoms by the bond.

For example : Water formation from Oxygen and Hydrogen is a chemical process :

2H_{2}+O_{2}\rightarrow 2H_{2}O

Original(old) bonds are :

H-H bond in H2 and O-O bonds in O2

In H2 = Hydrogen is joined to Hydrogen

IN O2 = Oxygen is joined to oxygen

New Bonds =

O-H bonds in water (H2O)

Oxygen is joined to hydrogen = New Bond formation

Hence,

<u>Chemical reactions do involve changes in the chemical bonds that join </u>

<u>atoms in compounds</u>

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3 years ago
If Jonathan went skateboarding from 4:00 PM to 4:30 PM and traveled 450 meters, what was his average speed?
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20 m Increase by five
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Energy transformation of a kettle?
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A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0
1 year ago
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