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alexgriva [62]
3 years ago
12

how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *

10^-4
Chemistry
1 answer:
Vitek1552 [10]3 years ago
7 0

Answer:

The number of moles =

Moles=4.97\times 10^{-8}

The number of molecules =

Molecules = 2.99\times 10^{16}

Explanation:

Volume of the sphere is given by :

V=\frac{4}{3}\pi r^{3}

here, r = radius of the sphere

radius=\frac{diameter}{2}

radius=\frac{6.0}{2}

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

V=\frac{4}{3}\pi 0.03^{3}

V=\frac{4}{3}\pi 2.7\times 10^{-5}

V=1.13\times 10^{-4}dm^{3}

V=1.13\times 10^{-4}L   (1 L = 1 dm3)

Now, concentration "C"=

C=4.4\times 10^{-4}moles/liter  

The concentration is given by the formula :

C=\frac{moles}{Volume(L)}

This is also written as,

Moles = C\times Volume

Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}

Moles=4.97\times 10^{-8}moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n"  mole of substance contain =( n x Na )

N_{a}=6.022\times 10^{23}

Molecules =

Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}

Molecules = 2.99\times 10^{16} molecules

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Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma
REY [17]

Answer:

34.7mL

Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

So now we need to go to HCl!

To do that we multiply by the molar coefficients in the chemical equation:

\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})

This leaves us with 2(0.0956) = 0.1912 mol HCl

Now we use the relationship M= moles / volume , to calculate our volume

Rearranging we get that V = moles / M

Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl

V= 0.0347 L

To change this to milliliters we multiply by 1000 so:

34.7 mL

7 0
4 years ago
When dissolved in water, hydrogen bromide (hbr) forms hydrobromic acid. determine the hydroxide ion concentration in a 4,500 ml
vovikov84 [41]

No of moles of HBr = 3.78g / 80.9 = 0.0467moles 
 Molarity = Number of moles/ volume. So we have 0.0467 / 4.5 = 0.0103M. 
 Kw = 1.0 x 10 -14 = [H3O+] [OH-] and [OH-] = 1.0 x 10^(-14) / [H3O+] 
 So [OH-] = 1.0 * 10^(-14) /1.03 *10^(-2) = 0.9703 * 10^(-12) M
7 0
3 years ago
3) What are the three types of sedimentary rocks?
astraxan [27]

Answer:

clastic, organic, and chemical

8 0
3 years ago
List the metric ladder from largest to smallest.
-Dominant- [34]

Answer:

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Explanation:

6 0
3 years ago
For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving a
Dvinal [7]

Answer: The correct option is e.

Explanation: In the given question it is asked to tell the element which has 'Sea of electrons' while bonding.

This means that the element should have extra electrons so that it can bond easily or this also means which element can easily donate its electron to form a bond.

  • Electronic configuration of Hydrogen = 1s^1

Hydrogen cannot loose electrons easily because the electron is present in 1s sub-shell which is the closest to the nucleus and hence will be tightly bonded to it.

  • Electronic configuration of Helium = 1s^2

This element will not loose electrons easily because the 1s sub-shell is fully filled and hence will not be available for bonding.

  • Electronic configuration of Sulfur = 3s^23p^4

This element lacks 2 electrons from attaining stable configurations. Hence, it will gain 2 electrons rather than donating for bonding.

  • Electronic configuration of Iodine = 5s^25p^5

This element lacks 1 electron to attain stable configuration. Hence, it will gain 1 electron rather than donating for bonding.

  • Electronic configuration of Lithium = 1s^22s^1

This element has an extra electron in its valence shell and can be easily lost in order to form bond.

Hence, the correct option is e.

4 0
4 years ago
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