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Sphinxa [80]
3 years ago
13

How many grams are in 1.76 x 10^23 atoms of iodine

Chemistry
1 answer:
Mariana [72]3 years ago
5 0

Answer:

\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

6.022*10^{23}

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

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

Multiply the given number of atoms by Avogadro's number.

1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I  }{1 \ mol \ I}

Flip the fraction so the atoms of iodine will cancel out.

1.76*10^{23} \ atoms \ I*\frac{  1 \ mol \ I}{6.022*10^{23} \ atoms \ I}

1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }

Multiply so the problem condenses into 1 fraction.

\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }

0.2922617071 \ mol \ I

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

  • Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply this fraction by the number of moles found above.

0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}

Multiply. The moles of iodine will cancel.

0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }

The 1 as a denominator is insignificant.

0.2922617071 *{ 126.9045 \ g\ I }

37.08932581 \ g \ I

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

37.08932581 \ g \ I

The 8 in the hundredth place tells us to round the 0 up to a 1.

\approx 37.1\ g \ I

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

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never [62]

Answer: B. 1:2

Explanation: Beryllium and chlorine forms a binary ionic compound. Ionic compound is formed when a metal loses its electrons to a receiving non metal. Beryllium (metal) has two valence electrons while chlorine (nonmetal) has seven valence electrons, and so a beryllium atom has to give out its two valence electrons to attain a duplet stable structure while a chlorine atom will gain one electron to attain its stable octet structure. In the reaction between beryllium and chlorine, two atoms of chlorine have to accept the two electrons from one beryllium atom to attain their stable octet structure.

The formula of the compound formed is BeCl2.

3 0
3 years ago
Describe the particle mode of matter. what does the brownian motion tells us about the particles in matter?give other evidence t
svetlana [45]

Answer:

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Explanation:

3 0
2 years ago
Convert 5.4g of NaCl to moles
sammy [17]

Answer:

.0924 moles of NaCl

Explanation:

So you know you have 5.4 g of NaCl and you need to know how many moles there are in this amount of NaCl

  1. You'll need to find the atomic mass of the compound NaCl to help you solve for moles
  • Sodium (Na) on the periodic table has a mass of 22.99
  • Chlorine (Cl) on the periodic table has a mass of 35.45

Add these two together----> 22.99 + 35.45 = 58.44

Now you can calculate for moles

<u>Written-out method:</u>

<u>5.4 grams of NaCl  |   1 mole of  NaCl              </u>

                                | 58.44 grams NaCl                = .0924 moles of NaCl

<u>Plug into calculator method:</u>

(5.4 g of NaCl/ 58.44g NaCl= .0925 moles)

7 0
2 years ago
Write a word equation for 2ZnS(s)+3O2(g) -&gt; 2ZnS(s) + 2SO3(g)
hjlf

2ZnS(s)+3O2(g) -> 2Zns(s) + 2SO3(g)

the above given equation is unbalanced as it contains 4 moles of sulphur in the output but in the input there are only two aoms of sulphur so to balance the equation we will write the equation as given under

balanced equation is

2ZnS(s)+3O2(g) -> 2Zn(s) + 2SO3(g)


In words:

When 2 moles of solid zinc sulfide reacts with 3 moles of oxygen gas gives 2 moles of solid zinc  and 2 moles of sulphur trioxide gas.

3 0
2 years ago
If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?
Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0
2 years ago
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