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2 years ago

Determine the quantity of bromine atoms in 18.0g of HgBr2 (using factor label method)

1 answer:

8 0

Considering the definition of molar mass and Avogadro's Number, the quantity of bromine atoms in 18.0g of HgBr₂ is 6.023×10²² atoms.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.

<h3>Molar mass of HgBr₂</h3>

In this case, you know the molar mass of the elements is:

Hg= 200 g/mole
Br= 79.9 g/mole

So, the molar mass of the compound HgBr₂ is calculated as:

HgBr₂= 200 g/mole + 2×79.9 g/mole

Solving:

HgBr₂= 359.98 g/mole

Definition of Avogadro's Number

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Quantity of bromine atoms in 18.0g of HgBr₂</h3>

Factor-label describes a technique to convert one quantity to another quantity. This method uses the fact that any number or expression can be multiplied by one without changing its value; this is, consists in multiply the value you are given, unit of measurement included, by the conversion factor in fraction form.

Finally, knowing that 2 moles of Br are present in 1 mole of HgBr₂, and considering the molar mass of HgBr₂ and the definition on Avogadro's number, the quantity of bromine atoms in 18.0g of HgBr₂ is calculated as:

$18 gr HgBr_{2}\frac{1 mole HgBr_{2}}{359.98 gHgBr_{2}} \frac{2 mole Br}{1 mole HgBr_{2} } \frac{6.023x10^{23}atoms Br }{1 mole Br} =6.023x10^{22}atoms$

In summary, the quantity of bromine atoms in 18.0g of HgBr₂ is 6.023×10²² atoms.

Learn more about

molar mass:

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Solnce55 [7]

The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

Let's consider the decomposition of ethanol on an alumina surface.

C₂H₅OH(g) ⇒ C₂H₄(g) + H₂O(g)

The plot of [A] vs time (t) resulted in a straight line, which indicates that the reaction follows zero-order kinetics.

The slope, 24.00 × 10²⁵ M/s, represents the rate constant, k.

<h3>What is zero-order kinetics?</h3>

It is a chemical reaction in which the rate of reaction is constant and independent of the concentration of the reacting substances

The rate law for zero-order kinetics is:

rate = 24.00 × 10²⁵ M/s

The integrated rate law for zero-order kinetics is:

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

Is the time for the amount of substance to decrease by half.

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, we can calculate the half-life [t(1/2)] using the following formula.

t(1/2) = [C₂H₅OH]₀ / 2 × k

t(1/2) = (1.25 × 10²² M) / 2 × (24.00 × 10²⁵ M/s) = 2.60 × 10⁻⁵ s

We can calculate the time required for all the 1.25 × 10²² M C₂H₅OH to decompose using the integrated rate law.

[C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

0 M = 1.25 × 10²² M - 24.00 × 10²⁵ M/s × t

t = 5.21 × 10⁻⁵ s

The rate constant for the decomposition of ethanol on an alumina surface is 24.00 × 10²⁵ M.

The rate law is: rate = 24.00 × 10²⁵ M/s

The integrated rate law is: [C₂H₅OH]t = [C₂H₅OH]₀ - 24.00 × 10²⁵ M/s × t

If the initial concentration of C₂H₅OH was 1.25 × 10²² M, the half-life is 2.60 × 10⁻⁵ s.

The time required for all the 1.25 × 10²² M C₂H₅OH to decompose is 5.21 × 10⁻⁵ s.

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The burning of ethane produces both co2 and H2O. if 400 ml of CO2 is produced at 30°C and 740 torr, what volume of water vapor w

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The volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.

If 400 ml of CO2 is produced at 30°C at 740 torr, then number of moles can be calculated as:

By using ideal gas equation:

P1V1 = N1R1T1

P1 = pressure = 740torr

V1 = 400 ml = volume of CO2

R = Gas constant = 8.314

T = 273+30 = 303 k

740×400 = N1×8.314×303

N1 = (740×400) /(8.314×303) =117.5.

Chemical equation

C2H6 ---- 2CO2 + 3H2O.

As we noticed from the equation that

2 moles of CO2 = 3 moles of H2O

1 moles of CO2 × 1 moles of H2O

Then N2 = 117.5 moles of CO2 = 3/2 × 117.5 moles of H2O

By using ideal gas equation:

P2V2 = N2RT2

V2 = 3/2 × 117.5 × 8.314 × 292/ 780

= 548.5ml.

Thus, we found that the volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.

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