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3 years ago

How many atoms of phosphorus are represented in 2Ca3(PO4)2? 2 4 8 16

Chemistry

2 answers:

Angelina_Jolie [31]3 years ago

6 0

We will have *1 2of2 atoms =p

irinina [24]3 years ago

5 0

Hello !

The number to the right of each element represents the number of the atoms of that element.


PO4 is taken twice so you will multiply 1 and 4 by 2.

Now we have 1*2=2 atoms of P


To the left,there is 2,so you will multiply all the elements by 2.It means that you'll multiply 2 (atoms of P) by 2 :


 2*2=4 atoms of phosphorus


ANSWER : 4

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Why did Dalton think it was important to use his system of symbols for the chemical elements?

77julia77 [94]

The answer is : A good luck :)

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3 years ago

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Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

c. Solution C : $[H_3O^+]=5.65\times 10^{-4}M$

$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

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3 years ago

34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i

nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

p(r) = $Ar(r^{2}e^{\frac{-r}{3a_{o}}}$

= $Ar^{3}e^{\frac{-r}{3a_{o}}}$

$\frac{dp(r)}{dr}$ = 0

$Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}}$ + $Ae^{\frac{-r}{3a_{o}}}(3r^{2})$ = 0

$\frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}$

= $3Ar^{2}e^{\frac{-r}{3a_{o}}}$

r = $9a_{o}$

Thus, we can conclude that most likely radius at which the electron would be found is $9a_{o}$ .

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3 years ago

Sulfuric acid dissolves aluminum metal according to the following reaction:

Fiesta28 [93]

Answer:

$m_{H_2SO_4}=81.7gH_2SO_4$

$m_{H_2}=1.67gH_2$

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

$2Al (s) + 3H_2SO_4 (aq)\rightarrow Al _2(SO4)_3 (aq) + 3H_2 (g)$

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

$m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4$

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

$m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2$

Best regards.

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3 years ago

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zubka84 [21]

Answer:

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Explanation:

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2 years ago