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andreyandreev [35.5K]
3 years ago
14

How many molecules are in one mole of a compound?

Chemistry
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:6.022 x 10^23 molecules

Explanation: Since one mole of any chemical compound always contains 6.022 x 10^23 molecules, you can calculate the number of molecules of any substance if you know its mass and its chemical formula.

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It is also used because it has the lowest vapor pressure and greatest tensile strength.
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Question 3 (5 points)
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Answer:

332.918g O2

Explanation:

I'm having some issues with the work however, your final answer should be 332.918g O2

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3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

6 0
3 years ago
Which term names a metamorphic change in rocks over a wide area from tectonic forces in Earth's crust and mantle? A. local metam
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A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting
n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

           = 3.09691250944

<em>From all the available options below:</em>

<em>a) 2.97 </em>

<em>b) 3.10 </em>

<em>c) 3.19 </em>

<em>d) 3.22 </em>

<em>e) 3.32</em>

The correct option is b.

4 0
3 years ago
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