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3 years ago

Why couple use a method of birth control

Chemistry

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

because it will control the birth of a child which can happen alot being overpopulated in the country or world. It also can mange their life and create a small and happy and healthy family.Also the child can get more love and support from their parents if they have less children and it wont be a taruma for the couple as well. because growing a child can be expensive time costly and really hard.

plz mark me as a brainlest a and thanks my answer

Explanation:

and btw i think the subject should be biology not chemestry

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Who is the Administrator of Brainly?? Hopefully, someone will answer (◍•ᴗ•◍)❤

sleet_krkn [62]

Answer:

Hey mate....

Explanation:

This is ur answer....

Kavisharma/ Haezel is the admin of brainly......

hope it helps,

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3 0

3 years ago

Read 2 more answers

Why are metals malleable,ductile,and good conductors

9966 [12]

The energy is transferred throughout the rest of the metal by the moving electrons. Metals are described as
malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond.

7 0

3 years ago

a rock is found to have a small amount of copper would you call the rock an ore of copper explain in 5-6 sentences

alexgriva [62]

Answer:

Answer: No. An ore is a rock having large amounts of a mineral. Minerals may or may not be commercially useful.

3 0

3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

The energy required to ionize boron is 801 kJ/mol. You may want to reference (Pages 93 - 98) Section 2.5 while completing this p

earnstyle [38]

Answer:

The frequency is $f = 2,01 * 10^{15} \ Hz$

Explanation:

From the question we are told that

The energy required to ionize boron is $E_b = 801 KJ/mol$

Generally the ionization energy of boron pre atom is mathematically represented as

$E_a = \frac{E_b}{N_A}$

Here $N_A$ is the Avogadro's constant with value $N_A = 6.022*10^{23}$

$E_a = \frac{801}{6.022*10^{23}}$

=> $E_a = 1.330 *10^{-18} \ J/atom$

Generally the energy required to liberate one electron from an atom is equivalent to the ionization energy per atom and this mathematically represented as

$E = hf = E_a$

=> $hf = E_a$

Here h is the Planks constant with value $h = 6.626 *10^{-34}$

$f = \frac{1.330 *10^{-18}}{ 6.626 *10^{-34}}$

=> $f = 2,01 * 10^{15} \ Hz$

8 0

2 years ago