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Dima020 [189]
3 years ago
12

Calculate the percent of lead in Pb (Co3)2

Chemistry
1 answer:
velikii [3]3 years ago
7 0
We will take that molar mass of Pb(CO3)2 represents the total mass of all particles in this compound, ie it has value 100%.

M(Pb(CO3)2) = Ar(Pb) + 2xAr(C) + 6xAr(O) = 207.2 + 2x12 + 6x16= 327.2 g/mol

M(Pb) = 207.2 g/mol

From the date above we can set the following ratio:

M(Pb(CO3)2) : M(Pb) = 100% : x

327.2 : 207.2 = 100 :x

x = 63.33% of Pb there is in <span>Pb(Co3)2</span>




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What is the mass of 13 gold atoms?
Ludmilka [50]

Answer:

uuuh 4.25191 × 10-21 grams????

Explanation:

8 0
3 years ago
The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.
Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions {\text{IO}_{3}}^{-}.

How many moles of iodine \text{I}_2 will be produced?

{\text{IO}_{3}}^{-} converts to \text{I}_2 in the first reaction. The coefficient in front of \text{I}_2 is three times the coefficient in front of {\text{IO}_{3}}^{-}. In other words, each mole of {\text{IO}_{3}}^{-} will produce three moles of \text{I}_2. 0.0020 moles of {\text{IO}_{3}}^{-} will convert to 0.0060 moles of \text{I}_2.

How many moles of thiosulfate ions {\text{S}_2\text{O}_3}^{2-} are required?

\text{I}_2 reacts with {\text{S}_2\text{O}_3}^{2-} in the second reaction. The coefficient in front of \text{I}_2 is twice the coefficient in front of {\text{S}_2\text{O}_3}^{2-}. How many moles of {\text{S}_2\text{O}_3}^{2-} does each mole of \text{I}_2 consume? Two. 0.0060 moles of \text{I}_2 will be produced. As a result, 2 \times 0.0060 = 0.0120 moles of {\text{S}_2\text{O}_3}^{2-} will be needed.

6 0
3 years ago
16. What is a pure substance formed when two or more different elementscombine called?A. compoundB. moleculeC. element
Lana71 [14]
<h2>Answer:</h2>

Compound. Option A is correct

<h2>Explanations:</h2>

When two or more different elements combines, a compound is formed. For instance, if carbon and oxygen combines, carbondioxide is formed according to the equation;

C+O_2\rightarrow CO_2

From the reaction, carbon and oxygen are the elements while carbondioxide is the compound. Hence we can conclude that a pure substance formed when two or more different elements combine called compound.

3 0
1 year ago
Estimate the standard internal energy of formation of liquid methyl acetate (methyl ethanoate, ch3cooch3) at 298 k from its stan
LenKa [72]

solution:

the reaction for formation of methyl acetate is\\CH_{3}OH+CH_{3}COOH----------->CH_{3}COOCH_{3}+H_{2}O\\standard internal energy \\\delta u=\delta H -\delta(pv)=\delta H -\delta n\times RT\\where \delta n=change in number of moles=0\\\delta u=\delta H\\=-442kj/mol

6 0
3 years ago
Read 2 more answers
The gas SF6 is used to trace air flows because it is non-toxic and can be detected selectively in air at a concentration of 1.0
melomori [17]

SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.

First, we will calculate the partial pressure of SF₆  using the following expression.

pSF_6 = P \times \frac{ppb}{10^{9} }

where,

  • pSF₆: partial pressure of SF₆
  • P: total pressure of air (we will assume it is 1 atm)
  • ppb: concentration of SF₆ in parts per billion

pSF_6 = P \times \frac{ppb}{10^{9} } = 1 atm \times \frac{1.0 ppb}{10^{9} } = 1.0 \times 10^{-9} atm

Then, we will convert 1.0 cm³ to L using the following conversion factors:

  • 1 cm³ = 1 mL
  • 1 L = 1000 mL

1.0 cm^{3} \times \frac{1mL}{1cm^{3}} \times \frac{1L}{1000 mL} = 1.0 \times 10^{-3} L

Next, we will convert 46 °C to Kelvin using the following expression.

K = \° C + 273.15 = 46 + 273.15 = 319 K

Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.

P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T}  = \frac{1.0 \times 10^{-9} atm  \times 1.0 \times 10^{-3} L}{(0.082 atm.L/mol.K) \times 319 K} = 3.8 \times 10^{-14} mol

Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.

3.8 \times 10^{-14} mol \times \frac{6.02 \times 10^{23} molecules  }{mol} = 2.3 \times 10^{10} molecules

SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.

You can learn more about partial pressure here: brainly.com/question/13199169

6 0
3 years ago
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