Helium on the Moon was found to be 0.420% 2He, 2.75% 3He, and 96.83% 4He. What is the average atomic mass of helium on the Moon?

The amount of heat required to raise the temperature of the 34.2 g sample of aluminum by 34 oC is 1043.52 J.

What is Temperature?

Temperature is a measure of the average kinetic energy of the particles in a substance. It is a physical property that determines the direction of heat flow between two objects or systems in contact with each other. Temperature is measured in degrees Celsius (°C) or Fahrenheit (°F), or in kelvin (K) in the International System of Units (SI).

The amount of heat (q) required to raise the temperature of a substance can be calculated using the formula:

q = m x c x ΔT

Where:

m = mass of the substance (in grams)

c = specific heat of the substance (in J/g oC)

ΔT = change in temperature (in oC)

Plugging in the values given:

m = 34.2 g

c = 0.9 J/g oC

ΔT = 34 oC

q = (34.2 g) x (0.9 J/g oC) x (34 oC)

q = 1043.52 J

Therefore, the amount of heat required to raise the temperature of the 34.2 g sample of aluminum by 34 oC is 1043.52 J.

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The median reaction at the second end point in the HCl and NaOH titration is: HCl + NaOH → NaCl + H2O

In a titration between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the reaction involved is the neutralization reaction between an acid and a base. The balanced equation for this reaction is:

HCl + NaOH → NaCl + H2O

In this reaction, one mole of HCl reacts with one mole of NaOH to form one mole of NaCl (sodium chloride) and one mole of water.

During the titration process, the reaction occurs gradually as the base is added to the acid solution.

The first end point of the titration is reached when the moles of HCl and NaOH are stoichiometrically equivalent, meaning they react in a 1:1 ratio. At this point, all the HCl has been neutralized by the NaOH, and no excess of either reagent remains.

However, if the titration is continued beyond the first end point, the reaction between HCl and NaOH can still occur, albeit in a different ratio.

The second end point refers to the point where the moles of NaOH added exceed the stoichiometrically required amount to neutralize the HCl completely. As a result, any excess NaOH added after the second end point reacts with the excess HCl in a 1:1 ratio.

Therefore, the median reaction at the second end point in the HCl and NaOH titration is:

HCl + NaOH → NaCl + H2O

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The primary porosity of a sediment or rock consists of the spaces between the grains that make up that material. The more tightly packed the grains are, the lower the porosity.

First, we write our reaction:

PbO + 2HNO3 ====> Pb(NO3)2 + H2O (it is already balanced)

We need the mass of Pb(NO3)2 to be produced, from 25 g of HNO3

Procedure: Stoichiometry

Data:

The molar mass of PbO=223.2 g/mol

The molar mass of HNO3 =63.01g/mol

The molar mass of Pb(NO3)2 =331.2g/mol

The molar mass of H2O=18.0g/mol

According to our reaction, we proceed like this:

2 x 63.01 g HNO3 ------- 1 x 331.2 g Pb(NO3)2

25 g HNO3 -------- X

X = (25 g x 1 x 331.2 g)/ 2 x 63.01 g

X = 65.70 g

Answer: 65.70 g Pb(NO3)2

Answer:

1. Options A and B

2. Options B and C

3.. B. Net ∆G = -16.7 KJ/mol; C. Net ∆G = -14.2 KJ/mol

Explanation:

1. The spontaneity of a chemical reaction depends on its standard free energy change, ∆G. If ∆G is negative, the reaction is favourable, but when it is positive, the reaction is unfavorable.

Therefore, since reaction A and B have ∆G to be positive, they are unfavorable

2. Coupling an unfavorable reaction to a favourable reaction can help the reaction to proceed in the forward direction as long as the net free energy change is negative.

Coupling reaction A and C, as well as reaction B and C will make the reactions to become favourable as net ∆G is negative in both instances.

3. A and C: net ∆G = 13.8 - 30.5 = -16.7 KJ/mol

B and C: net ∆G = 16.3- 30.5 = -14.2 KJ/mol

Nitrogen G is a colorless, tasteless gas that is the most plentiful element in Earth's atmosphere and is a constituent of all living matter of nitrogen.

Nitrogen is an essential gaseous element with an atomic number of 7 and an atomic weight of 14.0067. Nitrogen gas has two particles of Nitrogen therefore, the molecular formula of this gas is N2. It is a non-metallic element that settled Group 15

nitrogen. [ nī′trə-jən ] N. A nonmetallic element that constructs up about 78 percent of the atmosphere by capacity, Nitrogen is a nonmetal and the lightest member of group 15 of the periodic table

So we can conclude that Nitrogen is the chemical element with the symbol N and atomic number 7.

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The equation C2H4(g) + H2(g) + C2H6(g) → Caco3(s) + Cao(s) + CO2(g) shows an increase in entropy due to the formation of a gas as a product. Option A

In this equation, the reactants on the left-hand side consist of gases (C2H4 and H2), while the products on the right-hand side include a solid (Caco3) and a gas (CO2).

When a reaction involves a change from gaseous to solid or liquid states, there is typically a decrease in entropy because the particles become more ordered and constrained in the solid or liquid phase.

Conversely, when a reaction involves the formation of gases, there is generally an increase in entropy because gases have higher degrees of molecular motion and greater freedom of movement compared to solids or liquids.

In the given equation, the reactants include three gaseous compounds (C2H4, H2, and C2H6), and one of the products is a gas (CO2). Therefore, the overall entropy of the system increases during this reaction.

The equation Fe(s) + S(s) → FeS(s) does not show an increase in entropy. Both the reactants (Fe and S) and the product (FeS) are solids. Since solids have lower entropy compared to gases or liquids, the entropy of the system does not increase in this reaction. Option A

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Answer :

Explanation:

Here, we want to get the name of the given compound

From what we have, there are 4 carbon atoms in the principal chain

This means we are dealing with the suffix But-

Since there are no double bonds, it simply means that the compound is anm alkane

Thus, the correct name for the compound is

The n there serves to give the information that it is normal butane

Answer:

A. The reaction shifts to the right (products) and the concentrations of I and H₂ decrease.

Explanation:

If gas is removed from the system at equilibrium, the system will try to compensate for the loss by shifting the reaction in a direction that produces more gas molecules. This is known as Le Chatelier's principle, which states that a system at equilibrium will respond to a disturbance by shifting in a way that minimizes the effect of the disturbance.

In this case, since gas is being removed from the system, the reaction will shift to the side that produces more gas molecules. Looking at the balanced equation, we can see that 2HI(g) has a greater number of gas molecules compared to H₂(g) and I₂(g). Therefore, the system will shift to the right (products) to produce more HI(g) and reestablish equilibrium.

Answer:

First of all, it's KNO₃ not KNO.

Second, KNO₃ is neither an acid nor it is a base, infact, it is a salt and therefore it's neutral.

hope that helps...

The 2.9 grams would this oxygen (0₂) weigh in its liquid form.

What is oxygen ?

The sole sort of atom in oxygen makes it a chemical element, which is a substance. An oxygen atom has eight protons in its nucleus, as indicated by its atomic number of 8, which is represented by the letter O in its official chemical formula. At room temperature, oxygen is a gas that lacks all other flavors, smells, and colours. Molecular oxygen can be found in nature.

What is liquid ?

Unlike a solid, which is more rigid than a liquid, a liquid is a kind of substance with unique qualities. The opposite of a solid, which has a defined shape, is a liquid, which can flow. An alternative is that a liquid will take on the shape of the container it is held in.

Therefore, 2.9 grams would this oxygen (0₂) weigh in its liquid form.

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If your spectrophotometer can measure an absorbance up to 1.5 accurately, the maximum concentration of pNA that you can measure without diluting is 310 ug/mL.

A spectrophotometer counts photons to estimate the intensity of light spectra absorbed and transmitted by a sample. This gives information about the concentration of a compound in the sample.

This is determined by the wavelength at which the measurements are taken. The absorbance at 340 nm is commonly used for NADH because it is the best and cleanest peak in the absorbance curve. An absorbance of 1.5 equals 310 ug/mL.

Thus, the maximum concentration of pNA that you can measure without diluting is 310 ug/mL.

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The  low-spin octahedral complex [Co(CN)6]¯, is known to be one that all the 6 cyanide ligands that are said to be placed in an octahedral geometry and the 6 electrons from cyanide ligands that will tend to pair up with the 6 electrons from cobalt's 3d orbitals. So no unpaired electrons is seen this complex.

The complex [Co(CN)6]¯ is one that takes in a coordination number of 6, and cyanide (CN¯) may be a solid field ligand. Thus, the complex may be a low-spin octahedral complex.

The electronic arrangement of Co(II) particle is 3d⁷.

So, Cyanide (CN¯) could be a solid field ligand, which suggests it'll cause the 3d orbitals to match up some time recently moving onto the 4d orbitals, coming about within the arrangement of a low-spin complex.

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The layers of the atmosphere with the boundaries that are between them are shown in the image attached here.

The troposphere, which rises up to an altitude of roughly 10-15 km, is the layer that is closest to the Earth's surface. The majority of the Earth's air mass is contained here, which is also where weather happens.

The tropopause marks the transition from the troposphere to the stratosphere, the next layer. It designates the altitude at which the temperature stops dropping.

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The temperature in Celsius and kelvin is 50°C and 323.15 K.

The three most used temperature scales for usage in business, science, and daily life are Kelvin, Celsius, and Fahrenheit.

A scale in Kelvin is absolute. There are no degree symbols after its values, which begin at absolute zero.

Both Celsius and Fahrenheit are relative scales. The degree symbol is used to indicate Fahrenheit and Celsius temperatures.

Given the temperature is 122°F

Temperature in Celsius is -

⇒ C = (F - 32)/1.80

⇒ C = (122 - 32) / 1.80

⇒ C = 90 / 1.80

⇒ C = 50°C

Temperature in Kelvin is -

K = 5/9(F - 32) + 273.15

⇒ K = 5/9 (122 - 32) + 273.15

⇒ K = 323.15 K

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Answer:

its sodium hydroxide

Explanation:

Answer:

increase the number of gas particles

Explanation:

The general gas equation, commonly referred to as the ideal gas law, represents the state of a fictitious ideal gas through an equation. The molar mass of ethylene gas when the pressure is 755.3 mmHg, the temperature is 25.0 °C and it weighs 36.2449 g is 28.29 g/mol.

According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.

The ideal gas equation is:

PV = nRT

n = PV / RT

755.3 mmHg = 0.99 atm

25.0 °C = 298 K

R = 0.082 atm·L·mol⁻¹·K⁻¹

Mass of water = 142.3415 - 36.1235 = 106.218 g

Density = Mass / Volume

V = m / d = 106.218 / 0.9970 = 106.53 mL = 0.106 L

n = 0.99 × 0.106 / 0.082 × 298  = 0.00429 mol

Mass of ethylene gas = 36.2449 - 36.1235 = 0.1214 g

Molar mass =  0.1214 / 0.00429 = 28.29 g/mol

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1.34 x 10^{-3} is the concentration of NH4 .

What is pH ?

pH, a quantitative measure of the acidity or basicity of aqueous or other liquid solutions. Widely used in chemistry, biology, and agriculture, this term translates the concentration value of hydrogen ions. This typically ranges from about 1 to 10-14 gram equivalents per liter. Converted to a number between 0 and 14. At neutral (neither acidic nor alkaline), the concentration of hydrogen ions is 10−7 gram equivalents per liter, which corresponds to a pH of 7. Solutions below pH 7 are considered acidic. Solutions with a pH greater than 7 are considered basic or alkaline.

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Answer: 0.375L

Explanation: Using M1V1=M2V2, we can solve for V1.

12M*V1=1.5M*3.0L

V1 = \(\frac{1.5M*3.0L}{12M}\)

V1 = 0.375L.

0.375 L volume of the solution she needed to prepare 3.0 L of a 1.5 M HCl solution.

A homogenous mixture of one or more solutes that have been dissolved in a solvent is known as a solution. The material that a solute dissolves in to create a homogenous mixture is known as a solvent.

The component that dissolves in a solvent to create a homogenous mixture is known as a solute.

M₁ is the initial concentration of HCl solution = 12.0 M

V₁ is the volume of initial HCl solution that is needed

M₂ is the final concentration of HCl solution = 1.5 M

V₂  is the final volume of HCl solution to be prepared = 3.0 L

To find V₁, we get

V₁ = (M₂ V₂) / M₁

calculating the values, we will get:

V₁ = 1.5 M x 3.0 L / 12.0 M

= 0.375 L

Therefore, tech required to take 0.375 L of the 12.0 M HCl solution and dilute it to  3.0 L to make a 1.5 M HCl solution.

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Answer:

sodium perchorate is NaCLO4

calcium sulfite CaSO3

potassium hydroxide KOH

lithium nitrate LiNO3

Explanation:

sodium perchorate has one Sodium

one Chlorine and 4 Oxygen

calcium sulfite has one Calcium on sulfate and 3Oxygen

lithium nitrate has one Lithium oneNitrogen

3 Oxygen

The enthalpy change of the reaction C(diamond) → C(graphite) is -2.9 kJ.

The given information is ΔHrxn for the reaction C(diamond) → C(graphite) can be calculated with the given equations:Equations:  C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ  2 CO2(g) → 2 CO(g) + O2(g) ΔH = 566.0 kJ  C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ  2 CO(g) → C(graphite) + CO2(g) ΔH = −172.5 kJThe required reaction can be obtained by adding the equations (1) and (4), as follows:C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)Addition of the two equations (1) and (4) results in a reaction whose products are C(graphite) and CO2.

To get the final equation that involves only the required reactants and products, the equation (2) should be added, which consumes CO2 and produces O2, as shown below:C(diamond) + O2(g) → CO2(g)  ΔH = −395.4 kJ [eq. (1)]  2 CO(g) → C(graphite) + CO2(g)  ΔH = −172.5 kJ [eq. (4)]  2 CO2(g) → 2 CO(g) + O2(g)  ΔH = 566.0 kJ [eq. (2)]  C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)  ΔHrxn=ΣΔHf(products)−ΣΔHf(reactants)  ΔHrxn=[(3 mol CO2)(-393.5 kJ/mol) + (1 mol C(graphite))(0 kJ/mol)] − [(1 mol C(diamond))(0 kJ/mol) + (1 mol O2)(0 kJ/mol) + (2 mol CO(g))(−172.5 kJ/mol)] − [(2 mol CO2)(566.0 kJ/mol)]  ΔHrxn=−2.9 kJ.

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(a) the molecular formula of the compound is C₆H₁₂.

(b)  the molecular formula of the compound is NH₂Cl.

(a) Given the empirical formula CH₂ and a molar mass of 84 g/mol, we need to determine the molecular formula. To do so, we need to find the factor by which the empirical formula needs to be multiplied to achieve the given molar mass.

The empirical formula CH₂ has a molar mass of 14 g/mol (12 g/mol for carbon + 2 g/mol for hydrogen).

To find the factor, we divide the molar mass by the empirical formula mass:

Factor = (molar mass) / (empirical formula mass) = 84 g/mol / 14 g/mol = 6

Therefore, the molecular formula is obtained by multiplying the empirical formula by the factor:

CH₂ × 6 = C₆H₁₂

Thus, the molecular formula of the compound is C₆H₁₂.

(b) Given the empirical formula NH₂Cl and a molar mass of 51.5 g/mol, we follow a similar approach.

The empirical formula NH₂Cl has a molar mass of 51.5 g/mol (14 g/mol for nitrogen + 2 g/mol for each hydrogen + 35.5 g/mol for chlorine).

To find the factor, we divide the molar mass by the empirical formula mass:

Factor = (molar mass) / (empirical formula mass) = 51.5 g/mol / 51.5 g/mol = 1

Therefore, the molecular formula is the same as the empirical formula: NH₂Cl

Hence, the molecular formula of the compound is NH₂Cl.

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Answer:

Explanation:

Interesting question - especially if you are afflicted by something like Leukemia where it pays to know everything you can about Hemoglobin.

Molecules

1 mole of anything has 6.02 * 10^23 in this case molecules

6.4*10^-3 moles has x number of molecules

Set up the proportion

1/6.4*10^-3   =  6.02 * 10^23 molecule / x          Cross Multiply

x = 6.4*10^-3 * 6.02 * 10^23

x = 3.853*10^21 molecules

Mass Hemoglobin

1 mol of hemoglobin is 64456 grams

6.4*10^-3 moles = x grams                                  Cross multiply

x = 64456 * 6.4*10^-3

x = 412.5 grams of hemoglobin in this woman's body.

If you are American and think in pounds, then this is roughly a pound of hemoglobin.

Considering that you need hemoglobin to carry oxygen around to various organs, it is not an awful lot. That's the miracle of the human body.

Answer: hope this is hopeful

Explanation:  

1.So, potassium atom is bigger than sodium atom.

2.Sodium: The valence electron of sodium is in the 3s orbital. Potassium: The valence electron of potassium is in the 4s orbital. Atomic Size. Sodium: Sodium atom is

3. • While excess of sodium is harmful for us, low levels of potassium have also been found to be associated with certain disorders of lungs and heart.

The solubility of a substance, such as sugar, depends on several factors, including temperature, pressure, and the presence of other solutes. In this case, the student adds the same amount of sugar and tea bag to the same amount of hot water every day, but the temperature of the water could vary from day to day, affecting how much sugar dissolves.

According to the graph, the least amount of sugar dissolved on day 3, while the most sugar dissolved on day 5. The difference in temperature on these days could explain this variation. On day 3, the water may have been cooler, making it more difficult for the sugar to dissolve. On the other hand, on day 5, the water may have been hotter, which could have increased the solubility of the sugar.

To increase the amount of sugar that dissolves in the tea, the student could try using hotter water or stirring the sugar more vigorously to distribute it evenly throughout the water. Alternatively, the student could try adding the sugar gradually while stirring to give it more time to dissolve before adding more.

Answer:

I'm not 100% sure on this because mine hasn't been graded yet, but here are the answers I submitted.

Explanation:

Which day the least sugar dissolved and which day the most sugar dissolved:

Day 4 is when the least sugar dissolved, and Day 2 is when the most sugar dissolved.

What could have caused less sugar to dissolve on some days:

The temperature of the tea could have caused the sugar not to dissolve on some days.

What the student could do to his drink to make more sugar dissolve.

The student would need to add the sugar to the tea as soon as it's done boiling.  The Sugar will dissolve faster in a warmer tea due to more energy of movement.  

Be sure to stir the tea as you add the sugar.  Stirring the sugar into the tea speeds up the rate of dissolving by helping distribute the sugar particles throughout the tea.  

If the student were to use granulated sugar, those are smaller particles and have greater surface area.  Greater surface area allows for more contact between the tea and the sugar.

Answer:

what grade is this?

Answer:

D

Explanation:

The ratio of C6H12O6 (which will be referred to as "the carb") to oxygen is 1 to 6, so if we find an answer which has the same ratio, it should be chosen. A is 1:3
B is even worse with a ratio of the carb to oxygen of 1:2
C is the same as B, 1:2
D has a ratio of the carb to oxygen of 1:6, which is what we are looking for.