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slamgirl [31]
3 years ago
7

20 POINTS 2 QUESTIONS CAN YOU ANSWER THEM ???

Chemistry
2 answers:
Viktor [21]3 years ago
7 0
2. <span>1/8 = (1/2)³ </span>
<span>so it's 3 half lives.</span>
Reika [66]3 years ago
4 0

<u>Answer 1 :</u> The synthesis reaction is, N_2+3H_2\rightarrow 2NH_3

Explanation :

Synthesis reaction : It is a type of reaction in which the reactants present in elemental state react to give a single product.

(1) N_2+3H_2\rightarrow 2NH_3

This is a synthesis reaction.

(2) MgI_2+Br_2\rightarrow MgBr_2+I2

This is a single displacement reaction in which the most reactive element displaces the least reactive element.

(3) KClO_3\rightarrow KCl+O_2

This is a decomposition reaction in which larger reactant decomposes to give two or more products.

(4) HCl+NaOH\rightarrow NaCl+H_2O

This is a neutralization reaction in which an acid and a base react to give a salt and water as a product.

<u>Answer 2 : </u>The number of half-lives it take will be, 3

Formula used :

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = \frac{1}{8}a_o

a_o = Initial amount of the reactant

n = number of half lives

Putting values in above equation, we get:

\frac{1}{8}a_o=\frac{a_o}{2^n}

2^n=8=2^3

n = 3

Therefore, the number of half-lives it take will be, 3

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The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

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