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aev [14]
2 years ago
14

when a candle burns both physical and chemical changes take place at identify these changes give another example of a familiar p

rocess in which both the chemical and physical changes takes place. ​
Chemistry
1 answer:
sp2606 [1]2 years ago
8 0

Answer:

So, the melting of wax and vapourisation of melted wax are physical changes. Chemical Changes : The wax near flame burns and gives new substances like carbon dioxide, carbon soot, water vapour, heat and light. LPG is another example of a familiar process in which both the chemical and physical changes take place.

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Determine what’s wrong with these electronic configurations
Anika [276]

Answer:

Explanation:

19) it is 3d10 instead of 4d10

20) it is missing 3p6, and 4s2 before 3d5

21) Ra is not a noble gas

22) Cs is not a noble gas

7 0
2 years ago
In sodium chloride the ions are lined up in a __ patterns
adelina 88 [10]
Random or ionic bond pattern
6 0
3 years ago
Dissolving brass requires an oxidizing acid such as concentrated nitric acid. Nitrogen dioxide is produced as a byproduct in thi
polet [3.4K]

Answer:

                  Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

Explanation:

Step 1: Write down the chemical formulas of given substances,

                                    Copper Metal  =  Cu

                                    Nitric Acid  =  HNO₃

                                    Copper (II) Nitrate  =  Cu(NO₃)₂

                                    Nitrogen Dioxide  =  NO₂

                                    Water  =  H₂O

Step 2: Write down the unbalance Chemical equation,

                         Cu  +  HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 3: Balance Cu atoms on both sides;

The number of Cu atoms on both sides are same. Hence, there number will remain the same.

Step 4: Balance N atoms on both sides;

As there is 1 N atom on left hand side and 3 N atoms on right hand side, so we will multiply HNO₃ by 3 to balance N on both sides, hence,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  H₂O

Step 5: Balance O atoms on both sides;

As there are 9 O atom on left hand side and 9 O atoms on right hand side, so they are balance.

Step 6: Balance H atoms on both sides;

As there are 3 H atom on left hand side and 2 H atoms on right hand side, so we will multiply H₂O by 2 as,

                         Cu  +  3 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

By doing so the number of O atoms got imbalanced, so to balance O atoms again we will multiply HNO₃ by 4 as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  NO₂  +  2 H₂O

Now, The Cu and H atoms are balanced, and the O atoms are greater on left hand side and the N atoms are greater on right hand side, therefore we will multiply NO₂ by 2 to balance both N and O as,

                         Cu  +  4 HNO₃    →    Cu(NO₃)₂  +  2 NO₂  +  2 H₂O

7 0
3 years ago
Identify the acids and the bases in the chemical equation ​
agasfer [191]
<h3><u>Answer;</u></h3>

Acids;  HCl and HC5H5N+

Bases; C5H5N and Cl-

<h3><u>Explanation;</u></h3>
  • According to Bronsted-Lowry Theory an acid is a proton or hydrogen ion donor while a base is a proton or a hydrogen ion acceptor.
  • In this case,<u> both HCl and HC5H5N+ are acids</u> as <u>they are donors of hydrogen ions</u>. HCl is an acid to the forward reaction while HC5H5N+ is a acid to the reverse reaction.
  • On the other hand, <u>C5H5N and Cl- are bases</u>, <u>they are acceptors of hydrogen  ions</u>. Cl-  is a base in the reverse reaction while C5H5N is a base in the forward reaction.
8 0
3 years ago
The ionic radius for Na+ is 0.097 ηm and for Cl- is 0.181 ηm, the absolute value of the charge for each ion is 1.6x10-19 C, ε_o=
Vanyuwa [196]

Answer:

B = (2.953 × 10⁻⁹⁵) N.m⁹

Explanation:

At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.

That is,

Fa + Fr = 0

Fa = - Fr

Fa = (|q₁q₂|)/(4πε₀r²)

Fr = -B/(r^n) but n = 9

Fr = -B/r⁹

(|q₁q₂|)/(4πε₀r²) = (B/r⁹)

|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C

(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²

r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m

(k|q₁q₂|)/(r²) = (B/r⁹)

(k × |q₁q₂|) = (B/r⁷)

B = (k × |q₁q₂| × r⁷)

B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]

B = (2.953 × 10⁻⁹⁵) N.m⁹

6 0
2 years ago
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