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Gemiola [76]
2 years ago
15

Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t

his transformation is a change in the atomic radius of Fe—from RBCC = 0.12584 nm to RFCC = 0.12894 nm—and, in addition, a change in density (and volume). Compute the percentage volume change associated with this reaction. Indicate a decreasing volume by a negative number.
Chemistry
2 answers:
Alex2 years ago
8 0

Answer:

false thought ia ion of neon = clarity active

Explanation:

x = 81254 \: and \: y = 91284

fenix001 [56]2 years ago
3 0

Answer:

The percentage volume change associated with this reaction is 97.66%

Explanation:

Please, the solution is in the Word file attached

Download docx
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In Part B the given conditions were 1.00 mol of argon in a 0.500-L container at 27.0 ∘C . You identified that the ideal pressure
dimaraw [331]

Answer : The percent difference between the ideal and real gas is, 4.06 %.

Explanation : Given,

Ideal pressure (true value) = 49.3 atm

Real pressure (measured value) = 47.3 atm

The formula used to calculate percent difference is :

Percent difference = \frac{\text{True value - Measured value}}{\text{True value}} \times 100

Percent difference = \frac{(49.3- 47.3)atm}{49.3atm}\times 100

Percent difference = 4.06 %

Therefore, the percent difference between the ideal and real gas is, 4.06 %.

4 0
2 years ago
A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound
Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

          C:H:O  =     1/0.5  :   3/0.5  :   0.5/0.5

          C:H:O  =     2      :     6      :     1

The empirical formula of compound will be C₂H₆O

5 0
2 years ago
A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
2 years ago
What is the molarity of a solution that contains 0.500 mole of kno3 dissolved in 0.500-liter of solution?
belka [17]

Answer : The molarity of solution is, 1.00 M

Explanation : Given,

Moles of KNO_3 = 0.500 mol

Volume of solution = 0.500 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }KNO_3}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{0.500mol}{0.500L}=1.00mole/L=1.00M

Therefore, the molarity of solution is, 1.00 M

7 0
2 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
2 years ago
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