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SashulF [63]
3 years ago
6

You have a piece of gold jewelry weighing 9.35g. Its volume is 0.654cm 3 . Assume the metal is an alloy of gold and silver, whic

h have densities of 19.3g/cm 3 and 10.5g/cm 3 , respectively. Also assume that there is no change in volume when the pure metals are mixed. Calculate the percentage of gold (by mass) in the alloy.
Chemistry
1 answer:
Vesna [10]3 years ago
4 0

Answer:

57.54% is the percentage of gold by mass in the alloy.

Explanation:

The mass of the mixture is 9.35g. That is:

9.35g = Mass Gold + Mass Silver

Mass silver = 9.35g - Mass gold (1)

Now, the volume of the jewelry is the sum of volumes of gold and silver:

0.654cm³ = Mass Gold / 19.5g/cm³ + Mass silver / 10.5g/cm³ (2)

<em>Volume = Mass / Density.</em>

Replacing (1) in (2):

0.654cm³ = Mass Gold / 19.5g/cm³ + (9.35g - Mass gold) / 10.5g/cm³

0.654 = 0.051282 Mass Gold + 0.890476 - 0.095238 Mass gold

-0.236476 = -0.043956 Mass gold

Mass gold = 5.38g

And mass percent of gold in the alloy is:

5.38g / 9.35g * 100 =

<h3>57.54%</h3>

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(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed
soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃  → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • KClO₃: 2 moles  
  • KCl: 2 moles
  • O₂: 3 moles

The molar mass of the compounds is:

  • KClO₃: 122.45 g/mole
  • KCl: 74.45 g/mole
  • O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
  • KCl: 2 moles ×74.45 g/mole= 148.9 grams
  • O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that the gas occupies.
  • T is the temperature of the gas.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

  • P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
  • V= 165 mL= 0.165 L (being 1000 mL= 1 L)
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 \frac{atmL}{molK}× 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 \frac{atmL}{molK}× 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

4 0
2 years ago
The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74
Alex73 [517]
From  the  equation  above   the  reacting   ratio  of  KClO3   to  O2  is  2:3 therefore  the  number  of  moles  of  oxygen  produced  is  ( 4 x3)/2 =  6 moles  since   four  moles  of  KClO3  was  consumed
mass=relative  formula mass  x  number  of  moles
That  is   32g/mol x 6  moles  =192grams


8 0
3 years ago
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A certain first order reaction has a half-life of 54. 3 s. How long will it take (in s) for the reactant concentration to decrea
Maksim231197 [3]

Answer:

82.4 s

Explanation:

Find the NUMBEr of half lives...then multiply by 54.3

2.27 = 6.5 (1/2)^n

log (2.27/6.5) / log (1/2) = n = 1.52 half lives

1.52 * 54.3 = 82.4 s

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What do the coefficients located before certain molecules in each chemical equation represent?
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It represents the number of moles required of that molecule to balance the chemical equation, which means to have the reaction chemically happen and goes to completion.

For example:
CH4 + O2 --> H2O + CO2     that is not balanced

with the coefficients located
CH4 + 2O2 --> 2H2O + CO2    now with the coefficients the number of oxygen and hydrogen on each side are equal
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3 years ago
Which of the following has 62 neutrons, 46 protons, and 46 electrons?​
Vesna [10]
The answer is Silver (Ag)
6 0
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