An element is made up of _____.

The mass of 1.50 mol of C₅H₁₂ is calculated to be 108.09 g using the molar mass of the compound.

The molar mass of a compound is the total mass of all the atoms in the molecule. The compound C₅H₁₂ consists of 5 carbon atoms and 12 hydrogen atoms. Given that the atomic mass of carbon (C) is 12.01 g/mol and the atomic mass of hydrogen (H) is 1.01 g/mol, the molar mass of C₅H₁₂can be calculated as: (5 x 12.01 g/mol) + (12 x 1.01 g/mol) = 72.06 g/mol. Therefore, the mass of 1.50 mol C₅H₁₂ would be (1.50 mol x 72.06 g/mol) = 108.09 g.

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The mass of 1.50 mol of C5H12 (pentane) is 108.2247 grams. First, the molar mass of C5H12 is calculated by summing the atomic masses of carbon and hydrogen, then this molar mass is multiplied by the number of moles given (1.50 mol) to find the mass.

To calculate the mass of 1.50 mol of C5H12, we need to first determine its molar mass. The molar mass of a compound is the sum of the masses of all the atoms in one mole of that compound. For C5H12, which is pentane, the molar mass can be calculated as follows:

Now that we have the molar mass, we can find the mass for 1.50 mol:

This calculation assumes that the molar mass has been rounded to four significant figures to match the precision of 1.50 mol, thus the mass of 1.50 mol C5H12 is 108.2247 grams with four significant figures.

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Substances with weaker intermolecular forces evaporate faster due to higher vapor pressure and volatility.

Substances with weaker intermolecular forces evaporate faster than substances with stronger intermolecular forces.

Weak intermolecular forces lead to a higher vapor pressure and increased volatility in substances, causing them to evaporate at a faster rate.

Final answer:

The alcohol that would produce 3-pentanone upon oxidation is 3-pentanol, a secondary alcohol that transforms into the ketone when oxidized.

Explanation:

The alcohol that would produce 3-pentanone when oxidized is 3-pentanol. This type of reaction is an oxidation process where the alcohol, specifically a secondary alcohol, is oxidized to a ketone. The oxidation of alcohols depends on the class of the alcohol. In this case, 3-pentanol is a secondary alcohol, which means it has two alkyl groups attached to the carbon bearing the hydroxyl (-OH) group. When this alcohol is oxidized using appropriate oxidizing agents like chromic acid or pyridinium chlorochromate (PCC), it forms the corresponding ketone, which is 3-pentanone.

(1) The nonliving material with uniform structure has been Mineral. Thus, option A is correct.

(2) The rock formed by the cooling of volcanic eruption and lava has been the igneous rock.

(3) A sedimentary rock can become a metamorphic rock by exposure to high heat and great pressure.

The rock and minerals are the nonliving compositions of the nature that are rich source of many fuels, and natural gases.

The varying form of rocks are present in the nature, and are converted from one form to another.

(1) Rocks have been defined as the material with nonuniform structure and composition. Rocks are the rich mineral source. They are the uniform structure throughout the material.

Thus, the nonliving material with uniform structure has been Mineral. Thus, option A is correct.

(2) The rock has been converted from one form to another in the rock cycle. The sedimentary rocks are formed by the deposits from the sea. The breakdown of sedimentary rock result in the metamorphic rocks.

The metamorphic rocks are deep inside the crust.

The rock formed by the cooling of volcanic eruption and lava has been the igneous rock.

(3) The sedimentary rocks are converted to the metamorphic rocks by the application of high heat and pressure to the sediments.

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Answer : The moles of [tex]Fe(OH)_2[/tex] is, [tex]3.259\times 10^{-8}mole[/tex]

Explanation :

First we have to calculate the [tex]H^+[/tex] concentration.

[tex]pH=-\log [H^+][/tex]

[tex]10.37=-\log [H^+][/tex]

[tex][H^+]=4.26\times 10^{-11}M[/tex]

Now we have to calculate the [tex]OH^-[/tex] concentration.

[tex][H^+][OH^-]=K_w[/tex]

[tex]4.26\times 10^{-11}\times [OH^-]=1.0\times 10^{-14}[/tex]

[tex][OH^-]=2.35\times 10^{-4}M[/tex]

Now we have to calculate the [tex]Fe^{2+}[/tex] ion concentration.

The balanced equilibrium reaction will be:

[tex]Fe(OH)_2\rightleftharpoons Fe^{2+}+2OH^-[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Fe^{2+}][OH^-]^2[/tex]

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

[tex]1.8\times 10^{-15}=[Fe^{2+}]\times (2.35\times 10^{-4})^2[/tex]

[tex][Fe^{2+}]=3.259\times 10^{-8}M[/tex]

Now we have to calculate the moles of [tex]Fe(OH)_2[/tex].

[tex]\text{Moles of }Fe(OH)_2=\text{Molarity of }Fe(OH)_2\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }Fe(OH)_2=3.259\times 10^{-8}mole/L\times 1L=3.259\times 10^{-8}mole[/tex]

Therefore, the moles of [tex]Fe(OH)_2[/tex] is, [tex]3.259\times 10^{-8}mole[/tex]

The number of moles of Fe(OH)₂ dissolved in 1.0 liter of water buffered at pH = 10.37 is 3.29x10⁻⁸.

The reaction of Fe(OH)₂ solubility in water is:

Fe(OH)₂(s) ⇄ Fe²⁺(aq) + 2OH⁻(aq) (1)

The product of solubility constant for the above reaction is:

[tex] Ksp = [Fe^{2+}][OH^{-}]^{2} = 1.8\cdot 10^{-15} [/tex]    (2)

To find the number of moles of Fe(OH)₂ dissolved in the water, we need to find the concentrations of Fe²⁺ and OH⁻.

The OH⁻ concentration can be calculated from the pH:

[tex] pH + pOH = 14 [/tex]

[tex] pOH = 14 - pH = 14 - 10.37 = 3.63 [/tex]

[tex] pOH = -log([OH^{-}]) [/tex]

[tex] [OH^{-}] = 10^{-pOH} = 10^{-3.63} = 2.34 \cdot 10^{-4} M [/tex]

Now, we can find the Fe²⁺ concentration (eq 2)

[tex] [Fe^{2+}] = \frac{Ksp}{[OH^{-}]^{2}} = \frac{1.8\cdot 10^{-15}}{(2.34 \cdot 10^{-4} M)^{2}} = 3.29 \cdot 10^{-8} M [/tex]

The number of moles of Fe²⁺ is:

[tex] n_{Fe^{2+}} = [Fe^{2+}]*V = 3.29 \cdot 10^{-8} mol/L*1.0 L = 3.29 \cdot 10^{-8} \:moles [/tex]

From reaction (1), we have that 1 mol of Fe(OH)₂ produces 1 mol of Fe²⁺, so the number of moles of Fe(OH)₂ is:

[tex] n_{Fe(OH)_{2}} = n_{Fe^{2+}} = 3.29 \cdot 10^{-8} \:moles [/tex]

Therefore, 3.29x10⁻⁸ moles of Fe(OH)₂ will dissolve in 1 liter of water.

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The correct answer is 2

The grams of fat present is 12 grams.

Based on the given information,

• A slice of pizza contains 13 grams of protein, 29 grams of carbs.  

• The total calories present in the pizza is 280 kcal.  

Now there is a need to find the grams of fat present in the pizza,

Total calories = 280 kcal

Total calories of carbohydrates = 29 × 4 kcal (1 g of carb contain 4kcal of energy) = 116 kcal.

Total calories of protein = 13 × 4 kcal (1 g of protein contains 4kcal of energy) = 52 kcal.

Calories due to fat = (280 kcal) - (116 kcal + 52 kcal)

= 112 kcal

1 gram of fat contains 9kcal of energy. The amount of fat can be determined as,

Amount of fat = (112/9) g = 12.44 grams or 12 grams of fat.

Thus, the grams of fat present is 12 grams.

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Final answer:

To find the average atomic mass of weapons-grade uranium, calculate the weighted average based on the percentages of each isotope: 80% of 235.044 amu (uranium-235) and 20% of 238.051 amu (uranium-238), yielding an average atomic mass of 235.6454 amu.

Explanation:

To calculate the average atomic mass of weapons-grade uranium, which consists of 80% uranium-235 and 20% uranium-238, you multiply the mass of each isotope by its relative percentage (expressed as a decimal) and then sum the results. The atomic mass unit (amu) for uranium-235 is 235.044 and for uranium-238 is 238.051.

The average atomic mass can be calculated as follows:

(80% of uranium-235) 80/100 × 235.044 amu = 188.0352 amu

(20% of uranium-238) 20/100 × 238.051 amu = 47.6102 amu

Add these values together: 188.0352 amu + 47.6102 amu = 235.6454 amu

Therefore, the average atomic mass of weapons-grade uranium is 235.6454 amu.

The partial pressure of carbon dioxide (PCO2) in systemic arterial blood is approximately 40 mm Hg.

The partial pressure of a gas in a mixture, like blood, is a measure of the pressure that gas exerts independently, assuming ideal behavior. In the context of blood gases, PCO2 is a crucial parameter reflecting the concentration of dissolved carbon dioxide in the bloodstream.

Normal systemic arterial blood PCO2 levels typically range from 35 to 45 mm Hg. This range is maintained within a very narrow margin as part of the body's acid-base balance, regulated by the respiratory system and the kidneys.

A PCO2 value significantly higher than 45 mm Hg can indicate respiratory acidosis, a condition where the blood becomes too acidic due to excess carbon dioxide. On the other hand, a PCO2 value below 35 mm Hg may indicate respiratory alkalosis, where the blood becomes too alkaline due to insufficient carbon dioxide.

Therefore, a PCO2 of 100 mm Hg, 104 mm Hg, or 45 mm Hg does not represent the typical PCO2 in systemic arterial blood and would suggest an abnormal condition. A PCO2 of 40 mm Hg is within the normal range for systemic arterial blood PCO2.

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Answer: it must be 6.01 kJ.

Explanation:

Latent heat of freezing is the amount of heat released to convert 1 mole of liquid to 1 mole of solid at atmospheric pressure.

[tex]H_2O(l)\rightarrow H_2O(s)[/tex]

Given: latent heat of freezing is 6.01 kJ.

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to 1 mole of liquid at atmospheric pressure. If latent heat of freezing is 6.01 kJ, latent heat of fusion will be same but the heat will be absorbed.

[tex]H_2O(s)\rightarrow H_2O(l)[/tex]

Answer:

Volume in milliliters of the liquid should be used is 49 mL.

Explanation:

Mass of the ethylene glycol = 54.6 g = 0.0546 kg

Density = [tex]1114 kg/m^3[/tex]

Volume of ethylene gylcol = V

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]1114 kg/m^3=\frac{0.0546 kg}{V}[/tex]

[tex]V=\frac{0.0546 kg}{1114 kg/m^3}=4.90\times 10^{-5} m^3[/tex]

[tex]1 m^3=1000 L[/tex]

[tex]4.90\times 10^{-5} m^3=4.90\times 10^{-5}\times 1000=4.90\times 10^{-2} L=49 ml[/tex]

1 L = 1000 mL

Volume in milliliters of the liquid should be used is 49 mL.

Answer: ΔH°f is not zero

Explanation:

the change in enthalpy formation is zero when elements are pure.when they are zero it is mainly  because they are most basic,where you cannot form an element with an element.

solution:

These phase changes can be brought about by heating or cooling. ... This kind of phase change--liquid to gas--is called evaporation or vaporization. Water vapor can in turn be cooled to form liquid water. This kind of phase change--from gas to liquid--is referred to as condensation vaporization.  

Condensation is the change of the physical state of matter from gas phase into liquid phase, and is the reverse of vaporization. The word most often refers to the water cycle.

This is required answer