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2 years ago

What is the mass of 2.11 × 1024 sulfur atoms??

1 answer:

8 0

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We do as follows:

2.11x 10^24 atoms ( 1 mol / 6.022x10^23 atoms ) ( 32.06 g / 1 mol ) = 112.33 g sulfur

Hope this answers the question.

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Convert 4.77x10^24 molecules of SO2 to grams

tensa zangetsu [6.8K]

Those are the correct steps, young chemist. Don't be discouraged by an insane answer.

5 0

2 years ago

It takes energy to ionize any atom. It takes progressively more and more energy for each successive electron that is removed and

IrinaVladis [17]

Answer:

Removal of Third Electron

Explanation:

a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.

Ionization energy increases from bottom to top within a group, and increases from left to right within a period.

5 0

2 years ago

3) Calculate the percent by mass of 3.55 g NaCl dissolved in 88 g water.

kotykmax [81]

Answer:

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

Explanation:

When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.

A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:

$Percent by mass=\frac{mass of solute}{mass of solution} *100$

In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.

In this case:

mass of solute: 3.55 g
mass of solution: 3.55 g + 88 g= 91.55 g

Replacing:

$Percentbymass=\frac{3.55}{91.55}*100$

Percent by mass= 3.88%

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

5 0

2 years ago

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc = 3.74*10⁻³

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = 3.74*10⁻³

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago

A solution contains 25 grams of KNO3 dissolved in 200. grams of H2O. Which numerical setup can be used to calculate the percent

Furkat [3]

Answer:

% = 11.11%

Explanation:

To get the %m/m of any solution we should use the following expression:

%m/m = m solute / m solution * 100

we have the mass of solute, but not the mass of solution, however this can be calculated. solution is made using solute and solvent so:

m solution = 25 + 200 = 225 g

Now that we have the mass of solution, we can calculate the %:

%m/m = 25 / 225 * 100

%m/m = 11.11%

This is the %m/m of this solution

3 0

2 years ago

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