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lakkis
2 months ago
10

The volume of a gas is decreased from 100 liters at 173.0°C to 50 liters at a constant pressure. After the decrease in volume, w

hat is the new temperature of the gas?
Chemistry
1 answer:
eduard [2.7K]2 months ago
5 0
223.08 K First, we convert 173.0 °C to K: 173.0 °C + 273.16 = 446.16 K Using Charles' law, we can solve this problem: T₁V₂ = T₂V₁ Where: T₁ = 446.16 K V₂ = 50 L T₂ =? V₁ = 100 L Inputting the data: 446.16 K * 50 L = T₂ * 100 L Now let's calculate for T₂: T₂ = 223.08 K
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Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol
lions [2927]

1. The luminol stock solution has a molarity of 1.431 M.

2. In 2.00 L of the diluted spray, there are 0.12 moles of luminol.

3. The volume of the stock solution from Part A that contains the same number of moles present in the diluted solution from Part B is 83.86 ml.

Additional Information

Stoichiometry in Chemistry focuses on the quantitative aspects of chemical reactions, which includes calculations related to volume, mass, and the count of ions, molecules, and elements.

Key concepts in stoichiometry include:

  • 1. Relative atomic mass
  • 2. Relative molecular mass

This refers to the relative atomic mass of a molecule.

  • 3. Mole

A mole represents the number of particles in a substance equivalent to the number of atoms in 12 grams of carbon-12.

1 mole = 6.02 × 10²³ particles.

The quantity of moles can also be derived by dividing mass (in grams) by either the relative mass of an element or the relative mass of a molecule.

\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}

Luminol (C₈H₇N₃O₂) is utilized for detecting blood traces at crime scenes, due to its reaction with iron found in blood.

To prepare a luminol stock solution, 19.0 g of luminol is mixed into a total volume of 75.0 mL of water.

Thus, the molarity is calculated as:

  • 1. Moles of Luminol

- the relative molecular mass of Luminol:

= 8.C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams/mol.

Thus, we have:

moles = grams / relative molecular mass.

mole=\frac{19}{177}

moles = 0.1073.

2. Molarity (M)

M = moles / volume

M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}

M = 1.431.

  • b. The concentration of luminol in the spray bottle is 6.00 × 10⁻² M. Therefore, in a 2 L solution, the number of moles is:

moles = M × volume

moles = 6 × 10⁻² × 2

moles = 0.12.

  • c. The molarity of the stock solution (Part A) is 1.431 M.

The diluted solution (Part B) contains 0.12 moles of luminol.

To find the volume of the stock solution (Part A) that has the same moles as the diluted solution (Part B):

volume = moles / M

volume\:=\:\frac{0.12}{1.431}

volume = 0.08386 L = 83.86 mL.

Further Learning

moles of water you can generate

the amount of each atom in the chemical's formula

the proportion of hydrogen to oxygen atoms in 2 L of water

Keywords: mole, volume, molarity, Luminol, relative molecular mass

6 0
2 months ago
Read 2 more answers
A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes combustion, the e
castortr0y [3046]

Response:

9.606 g

Clarification:

Step 1: Write the balanced combustion equation

C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the moles for 11.27 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

11.27 g × (1 mol/18.02 g) = 0.6254 mol

Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O

The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol

Step 4: Calculate the mass for 0.2085 moles of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

0.2085 mol × 46.07 g/mol = 9.606 g

7 0
3 months ago
In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so
Alekssandra [3086]

Respuesta:

El oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Explicación:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

La oxidación se define como la pérdida de electrones. La oxidación provoca un aumento en el número de oxidación de un elemento.

Si se descompone esta reacción en sus mitades de reducción y oxidación

Se observa que, de los reactivos mencionados anteriormente,

H202 se convierte en H2O y O2

MnO4- + H+ se convierte en Mn2+ y H2O

El número de oxidación de Mn cambia de +7 en MnO4- a +2 en Mn2+ (lo que indica evidentemente una reducción)

El oxígeno en MnO4- no cambia su número de oxidación, ya que se mantiene en -2

El número de oxidación del oxígeno cambia de -1 en H2O2 a -2 en H2O y 0 en O2

El hidrógeno en H2O2 no cambia su número de oxidación, y su número de oxidación se mantiene en +1 tanto en H2O2 como en H2O.

Esto indica que H2O2 sufre tanto oxidación como reducción; más específicamente, el oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Espero que esto ayude

7 0
2 months ago
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