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rosijanka
1 month ago
8

A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. the partial pressure of neon at 300k is 8.87 k

pa. calculate (i) the volume and (ii) the total pressure of the mixture.
Chemistry
1 answer:
alisha [2.9K]1 month ago
5 0
To find the volume of the gas blend, we must first ascertain the mixture's overall pressure. This requires using the concept of partial pressure, which represents the pressure that each gas would exert if it occupied the container alone. The partial pressure equals the mole fraction multiplied by the total system pressure. This allows us to find the total pressure.

Pneon = xneonP
P = Pneon / xneon
P = 8.87 kPa / (225 / (225 + 320 + 175))
P = 8.87 kPa / 0.3125 = 28.384 kPa

Applying the ideal gas law, we can calculate volume:
PV = nRT
V = nRT / P

n = 225 mg ( 1 mmol / 20.18 mg) + 320 mg ( 1 mmol / 16.05 mg ) + 175 mg ( 1 mmol / 39.95 mg ) = 35.47 mg = 35467.0 g

V = 35467.0 (8.314) (300) / (28384) = 3116.68 m^3
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Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Ca2+, V5+, Br-
lorasvet [2795]

Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

  • The ionic compound must exhibit neutrality; its total charge should equal zero.
  • A binary ionic compound is formed from two distinct ions.

Ca²⁺ combines with either Br⁻ or S²⁻ to create binary ionic compounds.

  • CaS is created when Ca²⁺ pairs with S²⁻ resulting in the neutral binary ionic compound CaS.
  • CaBr₂ results from the combination of one mole of Ca²⁺ with two moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

V⁵⁺ can also unite with either Br⁻ or S²⁻ to produce binary ionic compounds.

  • V₂S₅ is formed when two moles of V⁵⁺ bond with five moles of S²⁻ yielding the neutral binary ionic compound V₂S₅.
  • VBr₅ is produced by combining one mole of V⁵⁺ with five moles of Br⁻ to form the neutral binary ionic compound VBr₅.

Thus, the empirical formulas for four binary ionic compounds that may be produced are: CaS, CaBr₂, VBr₅, and V₂S₅.

5 0
2 months ago
A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is
castortr0y [3046]

Answer: The overall width of a crystal measures 1.65 mm.

Explanation:

Horizontal distance separating the two molecules is 16.5 nm.

Width of the 10^5 molecules:16.5 nm\times 10^5=16.5\times 10^5 nm

1 nm=10^{-6} mm

The overall width of a crystal measured in millimeters=16.5\times 10^5\times 10^{-6} mm=1.65 mm

The overall width of a crystal is 1.65 mm.

7 0
1 month ago
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Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the
Tems11 [2777]

Answer:

THE MOLAR MASS OF XCL2 IS 400 g/mol

THE MOLAR MASS OF YCL2 IS 250 g/mol.

Explanation:

We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.

Number of moles = mass / molar mass

Molar mass = mass / number of moles.

For XCl2,

mass = 100 g

number of moles = 0.25 mol

Thus, molar mass = mass / number of moles

Molar mass = 100 g / 0.25 mol

Molar mass = 400 g/mol.

For YCl2,

mass = 125 g

number of moles = 0.50 mol

Molar mass = 125 g / 0.50 mol

Molar mass = 250 g/mol.

Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.

3 0
1 month ago
A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
VMariaS [2998]

Answer: Yes, there is sufficient sodium carbonate available.

Explanation:

In this scenario, according to the specified reaction:

Using stoichiometry, one can figure out the grams of sodium carbonate required to neutralize 1,665 g of sulfuric acid as outlined below:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)

Hence, the amount on hand is 2.0 kg, which leaves 0.2 kg as surplus, therefore:

A. Yes, there is sufficient sodium carbonate available.

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3Best regards.

5 0
24 days ago
What features of this model will help Armando answer the question?
lions [2927]

Answer:

The adjustable legs along with the sand table.

Note: The question is incomplete. The full question is presented below.

Using Models to Address Questions Regarding Systems

Armando’s class was examining images of rivers shaped by flowing water. Most rivers appeared wide and shallow, except for one, which was narrow and deep. The students theorized that this river's narrowness and depth are due to:

  • the steepness of the hill from which the water descends, or
  • the diminutive size of the sand grains the water flows through.

To explore the answer to the question of why this river is so narrow and deep, Armando created the model outlined below.

Explanation:

The model constructed by Armando will facilitate addressing the question due to specific features:

1. Adjustable leg - as one theory proposed by the class suggests that the steep hill affecting the water's path could be the reason for the river's dimensions, the adjustable legs are designed to be raised or lowered to alter the slope, allowing testing of this theory.

2. Sand table - this acts as the streambed. By modifying the size of the sand grains, students can examine the second hypothesis that smaller sand grains contribute to the river's narrowness and depth.

The outcomes of their experimentation will lead them to a conclusion.

5 0
1 month ago
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