Uranium is a metallic element that is used in nuclear reactors and nuclear weapons. the vast majority of uranium found on earth is in the form of uranium-238, an isotope with a mass number of 238, while the uranium that is used for nuclear reactors and weapons is uranium-235, an isotope with a mass number of 235. what is the atomic difference between uranium-235 and uranium-238? uranium-235 has three fewer protons than uranium-238. uranium-235 has three fewer neutrons than uranium-238. uranium-235 has three fewer electrons than uranium-238. uranium-235 has three fewer atoms than uranium-238.

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.

Answer: chromatography

Explanation:

Chromatography : It is a type of separation process in which the liquid or gas is separated into components at different rate. The principle of chromatography is that the substance of molecule are distributed between the liquid phases. It involves a stationary and a mobile phase.  Hence,chromatography is a laboratory technique that is used to  separate plant pigments.

Distillation : It is a type of separation process in which the mixture of liquid separated or evaporated at different temperatures.

Fractional distillation is separation of a liquid mixture into fractions which differ in boiling points by means of distillation by use of a fractionating column.

Filtration : It is a type of mechanical separation process in which the solid is separated from the liquid through the filter paper.

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

(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:

cfcl2

Explanation:edg

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.

The number of molecules present in 2.10 moles of CO₂ is 1.264×10²⁴ molecules. Therefore, option D is correct.

The mole is the amount of substance in a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is "mol".

In the International System of Units, the mole (symbol mol) is the unit of substance amount (SI). The amount of substance is a measurement of how many elementary entities of a given substance are present in an object.

From Avogadro's hypothesis as follows;

1 mole of any substance = 6.023×10²³ molecules.

This implies that:

1 mole of CO₂ = 6.023×10²³ molecules

With the above information, we can obtain the number of molecules in 2.10 moles of CO₂. This can be obtained as given below:

1 mole of CO₂ = 6.023×10²³ molecules

Then,

2.10 moles of CO₂ = 2.10 × 6.02×10²³

2.10 moles of CO₂ = 1.264×10²⁴ molecules

Thus, the number of molecules present in 2.10 moles of CO₂ is 1.264×10²⁴ molecules., option D is correct.

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The correct answer is 2

3 would be the answer to your question...

to answer your second question, it is D

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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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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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.

Answer. A. 183

B. 142

C. 0.03

D.11.3

E.24.5

Explanation:

The answer is 6mm

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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.