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Feliz [49]
3 years ago
7

Calculate the number of atoms in 2.5 moles of Si.

Chemistry
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

\boxed {\boxed {\sf About \ 1.5 * 10^{24} \ atoms \ Si}}

Explanation:

When converting from moles to atoms, we must use Avogadro's number. This number tells us there are 6.022 * 10²³ atoms in 1 mole. We can multiply this number by the number of moles.

First, we must set up Avogadro's number as a ratio.

\frac {6.022 \ * 10^{23} \ atoms \ Si }{1 \ mol \ Si}}

Next, multiply the number of moles by the ratio.

2.5 \ mol \ Si *\frac {6.022 \ * 10^{23} \ atoms \ Si }{1 \ mol \ Si}}

When we multiply, the moles of silicon will cancel.

2.5 * \frac {6.022 \ * 10^{23} \ atoms \ Si }{1}}

Since the denominator of the fraction is 1, we can cancel it out too.

2.5 *  {6.022 \ * 10^{23} \ atoms \ Si }

1.5055 * 10^{24} \ atoms \ Si

The original measurement (2.5 moles) has 2 significant figures (2 and 5). Therefore we must round to 2 sig figs. For this question, 2 sig figs is the tenth place.

The 0 in the hundredth place tells us to leave the 5 in the tenth place.

1.5 * 10^{24} \ atoms \ Si

There are about <u>1.5 * 10²⁴ atoms of silicon.</u>

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lutik1710 [3]

Answer:

4.62

Explanation:

Using the equation -log[H3O+] = pH,

-log(2.4 x 10^-5) = 4.62

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The attraction between molecules of methane in the liquid state is primarily due to
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Attraction between molecules of methane in liquid state is primarily due to "London dispersion force".

Explanation:

Methane is a non-polar and aprotic molecule. Hence there is no dipole moment in methane as well as no chance of hydrogen bonding formation by methane.

We know that all molecules contain electrons. Therefore transient dipole arises in every molecule due to revolution of electrons around nucleus in a non-circular orbit. Hence an weak intermolecular attraction force is always present in every molecule as a result of this which is termed as "London dispersion force".

So, attraction between molecules of methane in liquid state is primarily due to "London dispersion force".

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Assign oxidation states to each atom in each of the following species.?
DochEvi [55]
1. CuCl2
Cl: 2(-1)= 2-
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3 years ago
What is the reaction for CL^2 + 2 KBr —&gt; 2 KCL+Br^2 of 11 grams of potassium bromide?
mylen [45]

Answer:

All the amounts of reactants and products are:

KBr

          11.0g (given)

          0.0924 mol

Cl₂

           0.0462 mol

           3.28g

KCl

           0.0924 mol

           6.89 g

Br₂

         0.0462 mol

         7.39 g

         

Explanation:

<u>1. Balanced chemical equation (given)</u>

   Cl_2+2KBr\rightarrow 2KCl+Br_2

<u>2. Mole ratios</u>

     \dfrac{1molCl_2}{2molKBr}

      \dfrac{2molKCl}{2molKBr}

     \dfrac{1molBr_2}{2molKBr}

<u />

<u>3. Molar masses</u>

  • Molar mass Cl₂: 70.906g/mol
  • Molar mass KBr: 119.002 g/mol
  • Molar mass KCl: 74.5513 g/mol
  • Molar mass KBr: 159.808 g/mol

<u>4. Convert 11 grams of potassium bromide to moles:</u>

  • #moles = mass in grams / molar mass
  • #mol KBr = 11g / 119.002g/mol = 0.092435mol KBr

<u>5. Use the mole ratios to find the amounts of Cl₂, KCl, and Br₂</u>

a) Cl₂

       \dfrac{1molCl_2}{2molKBr}\times 0.092435molKBr=0.0462molCl_2

        0.0462175molCl_2\times 70.906g/molCl_2=3.28gCl_2

b) KCl

       \dfrac{2molKCl}{2molKBr}\times0.092435molKBr=0.0924molKCl

      0.092435molKCl\times 74.5513g/molKCl=6.89gKCl

c) Br₂

       

         \dfrac{1molBr_2}{2molKBr}\times 0.092435molKBr=0.0462molBr_2

         0.0462175molBr_2\times 159.808g/molBr_2=7.39gBr2

The final calculations are rounded to 3 sginificant figures.

3 0
3 years ago
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