An industrial chemist is studying a sample of an unknown metal. Describe two ways he could change the metal physically and two ways he could change the metal chemically to try to identify it.

Answer;

D, Xe (g)

Solution and explanation;

If 2g has a volume of 340ml.

Density is 1000/340*2 = 5.88g/litre.

-This rules out the two solids, choices 2) &3)

If 1 litre has mass 5.88g,

then 22.4 liters (volume at STP) has mass 5.88*22.4 = 131.8g/mol

molar mass Br2 = 80*2 = 160g/mol NO

molar mass Xe = 131.3g/mol = YES.

Answer is Xe

Final answer:

To determine the matter filling a closed container at STP given a 2.0 gram sample, you can use the ideal gas law equation to find the number of moles. Then, divide the number of moles by the volume and the gas constant to solve for the pressure. Compare the pressure obtained to known substances' vapor pressures at STP to identify the matter.

Explanation:

The question is asking for the specific type of matter that would uniformly fill a 340-milliliter, closed container at STP (Standard Temperature and Pressure) when given a 2.0-gram sample. To determine the matter, we can use the ideal gas law equation, PV = nRT, and rearrange it to solve for n, the number of moles. Then, we can use the molar mass of the substance to find its identity.

First, convert the volume from milliliters to liters by dividing it by 1000: 340 mL ÷ 1000 = 0.34 L. Next, convert the mass from grams to moles using the molar mass of the substance:

1.(Conversion factor) Given: 2.0 g sample, 1 mole = molar mass

2.(Calculation) Moles of substance = 2.0 g ÷ molar mass

Once you have the number of moles, divide it by the volume (in liters) and the universal gas constant (0.0821 L·atm/mol·K) and solve for the pressure:

1.(Ideal Gas Law) PV = nRT

2.(Substitution) P × 0.34 L = n × 0.0821 L·atm/mol·K

3.(Isolation) P = (n × 0.0821 L·atm/mol·K) ÷ 0.34 L

After solving for the pressure, compare it to known substances' vapor pressures at STP to determine the identity of the matter in the container.

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Answer:

Kp and Kc are 0.01266 and 145.17, respectively.

Explanation:

Please check document attached.

Answer:

Prone to storm damage and limited to particular areas of the ocean

Explanation:

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Even when the wave power is beneficent towards the environment care, it has a notable disadvantage when high energy waves are took by the energy stations by cause of electrical failures which lead to the station breakdown. On the other hand, there are hard-to-access areas in the ocean which restricts the enforcement of this type of energy.

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Final answer:

The compound Mn3(PO4)2 is correctly named as manganese(II) phosphate, based on the rule of naming ionic compounds and considering the oxidation state of the manganese ion.

Explanation:

The correct name of the compound Mn3(PO4)2 is manganese(II) phosphate. This is based on the rule of naming ionic compounds. With ionic compounds, the metal cation's oxidation state is represented in Roman numerals in parentheses if the metal can have more than one charge state.

In this case, the Mn3+ cation in Mn3(PO4)2 has an oxidation state of +2 since the phosphate ions PO4^3- balance out 3 Mn ions. Therefore, the compound is correctly named manganese(II) phosphate.

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The partial pressure of each component of the gas are:

1. The partial pressure of CH₄ is 276.48 KPa

2. The partial pressure of C₂H₆ is 27.34 KPa

3. The partial pressure of C₃H₈ is 3.38 KPa

We'll begin by calculating the mole fraction of each gas.

For CH₄:

Percentage of CH₄ = 90%

Total = 100%

Mole fraction CH₄ =?

Mole fraction = mole / total

Mole fraction CH₄ = 90 / 100

Mole fraction CH₄ = 0.9

For C₂H₆:

Percentage of C₂H₆ = 8.9%

Total = 100%

Mole fraction C₂H₆ =?

Mole fraction = mole / total

Mole fraction C₂H₆ = 8.9 / 100

Mole fraction C₂H₆ = 0.089

For C₃H₈:

Percentage of C₃H₈ = 1.1%

Total = 100%

Mole fraction C₃H₈ =?

Mole fraction = mole / total

Mole fraction C₃H₈ = 1.1 / 100

Mole fraction C₃H₈ = 0.011

Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:

1. Determination of the partial pressure of CH₄

Mole fraction CH₄ = 0.9

Total pressure = 307.2 KPa

Partial pressure of CH₄ =?

Partial pressure = mole fraction × Total pressure

Partial pressure of CH₄ = 0.9 × 307.2

Partial pressure of CH₄ = 276.48 KPa

2. Determination of the partial pressure of C₂H₆

Mole fraction C₂H₆ = 0.089

Total pressure = 307.2 KPa

Partial pressure of C₂H₆ =?

Partial pressure = mole fraction × Total pressure

Partial pressure of C₂H₆ = 0.089 × 307.2

Partial pressure of C₂H₆ = 27.34 KPa

3. Determination of the partial pressure of C₃H₈

Mole fraction C₃H₈ = 0.011

Total pressure = 307.2 KPa

Partial pressure of C₃H₈ =?

Partial pressure = mole fraction × Total pressure

Partial pressure of C₃H₈ = 0.011 × 307.2

Partial pressure of C₃H₈ = 3.38 KPa

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Answer:

276.48 atm → CH₄

27.3 atm → C₂H₆

3.38 atm → C₃H₈

Explanation:

Percentages of each gas, are the mole fraction

0.9 CH₄

0.089 C₂H₆

0.011 C₃H₈

Mole fraction = Partial pressure each gas/ Total pressure

0.9 = Partial pressure CH₄ / 307.2 kPa

307.2 kPa . 0.9 = 276.48 atm

0.089 = Partial pressure C₂H₆ / 307.2 kPa

307.2 kPa . 0.089 = 27.3 atm

0.011 = Partial pressure C₃H₈ / 307.2 kPa

307.2 kPa . 0.011 = 3.38 atm