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Lina20 [59]
3 years ago
5

Help this mirrorball, please​

Chemistry
2 answers:
grin007 [14]3 years ago
5 0

Answer:

Option B

Explanation:

Avagadro's hypothesis showed that at constant temperature and pressure equal volume of all gases contains equal no of molecules.

Avagadro's constant is known as 6.022×10^23

AnnyKZ [126]3 years ago
3 0

Which of the following represents Avogadro's hypothesis?

  • <em>Equal</em><em> </em><em>volumes</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>gases</em><em> </em><em>under</em><em> </em><em>same</em><em> </em><em>conditions</em><em> </em><em>of</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em>contain</em><em> </em><em>equal</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>molecules</em>

<u>Avogadro's</u><u> </u><u>hypothesis</u><u> </u><u>is</u><u> </u><u>an</u><u> </u><u>experimental</u><u> </u><u>gas</u><u> </u><u>law</u><u> </u><u>that</u><u> </u><u>relates</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>volume</u><u> </u><u>of</u><u> </u><u>gas</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>amount</u><u> </u><u>of</u><u> </u><u>gas</u><u> </u><u>present</u><u>.</u><u> </u><u>Its</u><u> </u><u>formula</u><u> </u><u>is</u><u> </u><u>:</u>

<u>\sf  \frac{V_1}{ n_{1} }  =  \frac{V_2}{ n_{2} }</u>

Where,

  • V1 is first volume
  • V2 is second volume
  • n1 is first amount of gas (in moles)
  • n2 is second amount of gas (in moles)
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swat32

Answer:

0.0983 M

Explanation:

First, we need to find the formulas of the reactants. Potassium forms the ion K⁺, and iodide is the ion I⁻, thus potassium iodide is KI. Silver forms the ion Ag⁺, and nitrate is the ion NO₃⁻, thus silver nitrate is AgNO₃. In the reaction, the cations will be replaced:

KI(aq) + AgNO₃(aq) → KNO₃(aq) + AgI(s)

AgI is an insoluble salt, so it will precipitate, and all nitrates are soluble, thus KNO₃ will be in the ionic form: K⁺ and NO₃⁻. 1 mol of KNO₃ = 1 mol of K⁺.

The molar mass of KI is 166 g/mol, thus the number of moles that is added is:

nKI = mass/molar mass

nKI = 5.71/166 = 0.0344 mol

And the number of moles of AgNO₃ is given as 64mM = 0.064 mol. Because the stoichiometry is 1:1, AgNO₃ is in excess, thus, all the KI will react and form 0.0344 mol of KNO₃. So, nK⁺ = 0.0344 mol. The molarity is the number of moles divided by the volume (350 mL = 0.350 L):

0.0344/0.350 = 0.0983 M

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How many molecules are in 3.6 grams of NaCl? Question options:
Levart [38]

Answer:

\boxed {\boxed {\sf 3.7 * 10^{22} \ molecules \ NaCl}}

Explanation:

We are asked to find how many molecules are in 3.6 grams of sodium chloride.

<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

  • Na: 22.9897693 g/mol
  • Cl: 35.45 g/mol

There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

  • NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

3.6 \ g \ NaCl *\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}

Flip the ratio so the units of grams of sodium chloride cancel.

3.6 \ g \ NaCl *\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}

3.6  *\frac { 1 \ mol \ NaCl}{58.4397693}

\frac { 3.6}{58.4397693} \ mol \ NaCl

0.06160188589 \ mol \ NaCl

<h3>2. Convert Moles to Molecules </h3>

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

\frac {6.022 \times 10^{23} \ molecules \ NaCl}{ 1 \ mol \ NaCl}

Multiply by the number of moles we calculated.

0.06160188589 \ mol \ NaCl * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 \ mol \ NaCl}

The units of moles of sodium chloride cancel.

0.06160188589 * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 }

3.70966557*10^{22} \ molecules \ NaCl

<h3>3. Round </h3>

The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

3.7 * 10^{22} \ molecules \ NaCl

There are 3.7 * 10^{22} \ molecules \ NaCl in 3.6 grams and the correct answer is choice D.

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3 years ago
Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. I
Aneli [31]

Answer : The number of iron atoms present in each red blood cell are, 1.077\times 10^9

Explanation :

First we have to calculate the moles of iron.

\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles

Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains 6.022\times 10^{23} number of iron atoms

So, 0.0519 mole of iron contains 0.0519\times 6.022\times 10^{23}=3.125\times 10^{22} number of iron atoms

Now we have to calculate the number of iron atoms are present in each red blood cell.

Number of iron atoms are present in each red blood cell = \frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}

Number of iron atoms are present in each red blood cell = \frac{3.125\times 10^{22}}{2.90\times 10^{13}}

Number of iron atoms are present in each red blood cell = 1.077\times 10^9

Therefore, the number of iron atoms present in each red blood cell are, 1.077\times 10^9

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3 years ago
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