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tensa zangetsu [6.8K]
3 years ago
7

How many formula units is 213 g of PbCl2?

Chemistry
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

4.61 x 10^23

Explanation: Convert 213g of PbCl2 to mol, which is 0.7658950184462004, multiply 0.765 with 6.022 x 10^23.

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A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti
Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

Molarity = \frac{mol}{L}

Molarity of crystal violet = 3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = 3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = \frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}

Final concentration of CV = 0.0027 M

Molarity of NaOH= \frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}

NaOH is a strong base and dissociates completely as follows.

NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

8 0
3 years ago
PLEASE HELP!!!!!!!!! 12 POINTS!!!!!!!
Nimfa-mama [501]
Ocean-dwelling I would say.
4 0
3 years ago
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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
2 years ago
Determine the empirical formula of a
k0ka [10]

Answer:

AuCl

Explanation:

Given parameters:

Mass of Gold  = 2.6444g

Mass of Chlorine  = 0.476g

Unknown:

Empirical formula  = ?

Solution:

Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.

Elements                                     Au                                             Cl

Mass                                         2.6444                                     0.476

Molar mass                                 197                                          35.5

Number of moles                  2.6444/197                                 0.476/35.5

                                                 0.013                                           0.013

Divide by the

smallest                                 0.013/0.013                                 0.013/0.013

                                                       1                                                   1

The empirical formula of the compound is AuCl

3 0
2 years ago
Plato answers . Which field will she use
Ira Lisetskai [31]

Answer:

A. biology

Explanation:

biology is the human body

6 0
3 years ago
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