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Neporo4naja [7]
3 years ago
10

Why are flowers brightly colored?

Chemistry
1 answer:
mafiozo [28]3 years ago
8 0

Answer: In order to help the plants reproduce, flowers that are bright in color are meant to attract birds, bees and other insects. In the genetics of a flower, bright or dull colors are set in

Explanation:

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Write the ionic charges (such as Ca2+) and chemical formulas and fill-in the table below.
Vikentia [17]

1) Lithium and fluorine:

Ionic charges: lithium cation Li⁺ and fluorine anion F⁻.

Chemical formula LiF.

In ionic salt lithium fluoride (LiF), fluorine has electronegativity approximately χ = 4 and lithium χ = 1 (Δχ = 4 - 1; Δχ = 3).

Fluorine attracts electron and it has negative charge and lithium has positive charge.

2) Beryllium and oxygen:

Ionic charges cation Be²⁺ and anion O²⁻.

Chemical formula is BeO.

Beryllium is metal from group 2 and oxygen is nonmetal from group 16.

Electron configuration of beryllium: ₄Be: 1s² 2s², it has two valence electrons in 2s orbital.

Beryllium lose two electrons and to gain electron configuration as noble gas helium (He).

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen gain two valence electron to form anion with stable electron configuration as noble gas neon (atomic number 10).

3) Magnesium and fluorine:  

Ionic charges cation Mg²⁺ and anion F⁻.

Chemical formula is MgF₂.

Magnesium fluoride (MgF₂) is salt, ionic compound.  

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).  

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

Fluorine (F) is nonmetal with greatest electronegativity, which means it easily gain electrons.  

Fluorine has atomic number 9, which means it has 9 protons and 9 electrons. It gain one electron to form fluorine anion (F⁻) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.  

4) Aluminum and chlorine:  

Ionic charges cation Al³⁺ and anion Cl⁻.

Chemical formula is AlCl₃.

The right name for AlCl₃ is aluminium chloride.

Aluminium chloride is a salt with ionic bonds.

Aluminium (metal from group 13) has oxidation number +3 and chlorine (nonmetal from group 17) has oxidation number -1, chemical compound has neutral charge (+3 + 3 · (-1) = 0).

5) Beryllium and nitrogen:  

Ionic charges cation Be²⁺ and anion N³⁻.

Chemical formula is Be₃N₂.

Atomic number of nitrogen is 7, it has 7 protons and 7 electrons.

Electron configuration of nitrogen atom: ₇N 1s² 2s² 2p³.

Nitrogen gain three electrons to form anion with stable electron configuration as noble gas neon (atomic number 10).

4 0
3 years ago
Please help on this one?
quester [9]

Answer:

the answer is= NUCLEAR FISSION, NUCLEAR FUSSION, RADIOACTIVE DECAY.

7 0
4 years ago
Read 2 more answers
Assume that the partial pressure of sulfur dioxide, pso2, is equal to the partial pressure of dihydrogen sulfide, ph2s, and ther
Alex Ar [27]
<span>Kp = PH2O^2/(PH2O)^2 To calculate SO2 is to have a Kp for the reaction let x=PSO2=PH2S Kp= (28/760atm)^2/x^3 x=cube root(0.03684)/Kp=PSO2</span>
8 0
4 years ago
A voltaic cell consists of a Zn&gt;Zn2+ half-cell and a Ni&gt;Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn
nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of Ni^{2+} has fallen to 0.500 M is, 0.52 V

(c) The concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

E^0_{[Ni^{2+}/Ni]}=-0.23V

E^0_{[Zn^{2+}/Zn]}=-0.76V

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : Zn\rightarrow Zn^{2+}+2e^-     E^0_{[Zn^{2+}/Zn]}=-0.76V

Reaction at cathode (reduction) : Ni^{2+}+2e^-\rightarrow Ni     E^0_{[Ni^{2+}/Ni]}=-0.23V

The balanced cell reaction will be,  

Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}

E^o=(-0.23V)-(-0.76V)=0.53V

(a) Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}

E_{cell}=0.49V

(b) Now we have to calculate the cell potential when the concentration of Ni^{2+} has fallen to 0.500 M.

New concentration of Ni^{2+} = 1.50 - x = 0.500

x = 1 M

New concentration of Zn^{2+} = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}

Now put all the given values in the above equation, we get:

E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}

E_{cell}=0.52V

(c) Now we have to calculate the concentrations of Ni^{2+} and Zn^{2+} when the cell potential falls to 0.45 V.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}

Now put all the given values in the above equation, we get:

0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}

x=1.49M

The concentration of Ni^{2+} = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of Zn^{2+} = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0
4 years ago
Name three elements that have been added to the periodic table since mendeleev's time
dolphi86 [110]
Lanthamides, actinides, noble gases
7 0
3 years ago
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