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Alja [10]
2 years ago
7

5. A compound with a formula of lead (Pb) and sulfur (S) was determined using the method in this experiment. A sample of Pb was

weighed into a porcelain crucible and covered with finely powdered S. Then the crucible was covered and heated to allow the Pb and S to react. After additional heating, all the unreacted S was burned off. The crucible was then cooled, weighed, and finally showed the following data: Mass of crucible and cover, 25.429 grams Mass of crucible, cover, and Pb, g 29.465 grams Mass of crucible, cover, and compound formed, g 30.107 grams What is the mass of S that reacted?
Chemistry
1 answer:
Airida [17]2 years ago
7 0

Answer:

The mass of S that reacted is 0.642g

Explanation:

In this case we have

⇒ mass of crucible and cover, 25.429g

⇒ mass of crucible, cover, and Pb, 29.465g

⇒ mass of crucible, cover and commass of crucible and coverpound formed, 30.107g

The mass of Pb reacted = mass of crucible, cover, and Pb - mass of crucible and cover = 29.465 - 25.429 = 4.036g

The mass of compound formed = mass of crucible, cover and commass of crucible and coverpound formed  - mass of crucible and cover = 30.107 - 25.429g = 4.678g

The mass of S reacted = Mass of compound formed - mass of Pb reacted = 4.678 -4.036 = 0.642g

The mass of S that reacted is 0.642g

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Answer:

The answer to your question is below

Explanation:

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1.- Mass                          it does not depend on the type of compound

2.- Conductivity      -conduct electricity               - do not conduct electricity

                                  in solution.

3.- Color                  - Shiny                                     - opaque

4.- Melting point     - high                                       - lower than ionic compounds

5.- Boiling point     - high                                        - lower than ionic compounds

6.- flammability      - not flammable                       - flammable

6 0
2 years ago
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A sample of cacl2⋅2h2o/k2c2o4⋅h2o solid salt mixture is dissolved in ~150 ml de-ionized h2o. the oven dried precipitate has a ma
Paraphin [41]

We are given that the balanced chemical reaction is:

cacl2⋅2h2o(aq) + k2c2o4⋅h2o(aq) ---> cac2o4⋅h2o(s) + 2kcl(aq) + 2h2o(l)

We known that the product was oven dried, therefore the mass of 0.333 g pertains only to that of the substance cac2o4⋅h2o(s). So what we will do first is to convert this into moles by dividing the mass with the molar mass. The molar mass of cac2o4⋅h2o(s) is molar mass of cac2o4 plus the molar mass of h2o.

molar mass cac2o4⋅h2o(s) = 128.10 + 18 = 146.10 g /mole

moles cac2o4⋅h2o(s) = 0.333 / 146.10 = 2.28 x 10^-3 moles

Looking at the balanced chemical reaction, the ratio of cac2o4⋅h2o(s) and k2c2o4⋅h2o(aq) is 1:1, therefore:

moles k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles

Converting this to mass:

mass k2c2o4⋅h2o(aq) = 2.28 x 10^-3 moles (184.24 g /mol) = 0.419931006 g

 

Therefore:

The mass of k2c2o4⋅<span>h2o(aq) in the salt mixture is about 0.420 g</span>

3 0
2 years ago
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a scientists finds an organism that cannot move. it has many cells, produces spores, and gets food from its environment. in whic
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The correct answer is : The Organism belongs in Kingdom Fungi


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3 0
2 years ago
What is the pH of a 0.75 M HNO3 solution
Sauron [17]
Hello!

The dissociation reaction of HNO₃ is the following:

HNO₃ → H⁺ + NO₃⁻

This is a strong acid, so the concentration of HNO₃ would be the same as the concentration of H⁺. The formula for pH is the following:

pH=-log([H_3O^{+}])=-log(0,75M)=0,12

So, the pH would be 0,12

Have a nice day!
4 0
2 years ago
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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
2 years ago
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