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3 years ago

Investigate: Press Reset. Start with 2.000 moles of sulfur, then press Start. How many atoms do you think this amount represents

Chemistry

1 answer:

Scilla [17]3 years ago

6 0

Answer:

1.204428 * 10^24 atoms

Explanation:

Number of moles = 2 mol

Number of atoms = ?

The relationship between moles and atoms is given by the avogadro's umber. This is the number of units in one mole of a substance. The units can be atoms, ions etc In this case it is atoms. The number is equal to 6.02214076 * 10^23

This means;

1 mol = 6.02214076 * 10^23

2 mol = x

Upon solving for x,

x = 2 * 6.02214076 * 10^23

x = 12.04428 * 10^23

x = 1.204428 * 10^24 atoms

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<span>1. </span>To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

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3 years ago

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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

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

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Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

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

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

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

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

Moles of lead (II) iodide = 0.26 moles

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Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

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3 years ago

Heterotrophic bacteria obtain food by

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4 years ago

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Answer:

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Alex17521 [72]

Explanation:

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2 years ago

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