What is the maximum mass, in grams, of NH3 that can be produced by the reaction of 1. 0 g of N2 and 1. 0 g of H2 using the reaction below? N2 H2→NH3 (not balanced).

In the ice crystal, there are four more water molecules.Each water molecule has the ability to create four hydrogen bonds.

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Answer:

Neptune and Saturn although Neptune isn't on there so just Saturn.

hope it'll help ya out!

The enthalpy change for a phase change can be determined using calorimetric equation. The enthalpy change of 500 ml or 500 g of water for a temperature change of 25 to 62.8°C is 79002 J.

Calorimetry is an analytical technique used to determine the enthalpy change accompanied in a chemical reaction or physical change. The calorimetric equation connecting the heat energy q, with the mass m, specific heat c and the temperature change is given as follows:

q = m c ΔT.

Given that m = 500 ml = 500 g of water.

c for water = 4.18 J/°C g

ΔT = 62.8 - 25 = 37.8 °C

Then, change in enthalpy ΔH = 500 g × 37.8 °C ×4.18 J/°C g = 79002 J or 79 kJ.

Therefore, the enthalpy change for the given temperature change of water is 79 kJ.

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62.5 grams of water can be heated from 20.0°C to 75.0°C using 12500.0 J of energy.

To calculate the amount of water that can be heated from 20.0°C to 75.0°C using 12500.0 J of energy, we can use the following formula: Q = m * c * ΔT where Q is the amount of heat energy absorbed by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.

We know that Q is equal to 12500.0 J, c is equal to 4.18 J/g°C, and ΔT is equal to 75.0°C - 20.0°C = 55.0°C. Substituting these values into the formula, we get:

12500.0 J = m * 4.18 J/g°C * 55.0°C

Solving for m, we get:

m = 12500.0 J / (4.18 J/g°C * 55.0°C) = 62.5 g

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Litmus is an example of an indicator, a substance that changes color depending on its pH (pH is a measure of the concentration of protons, or H+ ions). In the Titration Gizmo™, you will use indicators to show how acids are neutralized by bases, and vice versa.

To begin, check that 1.00 M NaOH is selected for the Burette, Mystery HBr is selected for the Flask, and Bromthymol blue is selected for the Indicator.

Calculate: Concentration is measured by molarity (M), or moles per liter. Brackets are also used to symbolize molarity. For example, if 0.6 moles of HNO3 are dissolved in a liter of water, you would say [HNO3] = 0.6 M.

Because HNO3 is a strong acid, it dissociates almost completely in water. That means the concentration of H+ is very nearly equal to that of HNO3.What is [H+] if [HNO3] is 0.01 M? 0.01 M

The pH of a solution is equal to the negative log of H+ concentration: pH = –log[H+]

Describe: The equation for the reaction of nitric acid (HNO3) and sodium hydroxide (NaOH) is shown on the bottom right of the Gizmo.

Measure: A titration can be used to determine the concentration of an acid or base by measuring the amount of a solution with a known concentration, called the titrant, which reacts completely with a solution of unknown concentration, called the analyte. The point at which this occurs is called the equivalence point.

Explain: A titration curve is a graph of pH vs. volume of titrant. The graph at right shows a typical titration curve for the titration of a strong acid by a strong base. (A strong base is one that has relatively high dissociation in water.)

According to theBrønsted-Lowry definition, an acid is a substance that is capable of donating a proton to another substance. A base is a substance that accepts protons. When an acid and a base are combined, the acid is neutralized as the base accepts the protons produced by the acid.One way to determine if a solution is acidic or basic is to use litmus paper, as shown above. There are two types of litmus papers: red and blue.How does litmus paper indicate an acid? Both strips turn red.

The given statement "  The point at which indicator undergoes color change is called end point titration." True.

The endpoint of a titration is the point at which the indicator being used undergoes a color change, indicating that the reaction between the analyte and the titrant is complete. The choice of indicator depends on the nature of the reaction being studied and the pH range in which the reaction occurs. The indicator is selected such that its color changes at the pH at which the reaction is complete. For example, phenolphthalein is commonly used as an indicator in acid-base titrations because it changes from colorless to pink in the presence of a base, indicating the endpoint of the titration.

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The pH of a 0.5 M solution of sodium bisulfate (\(NAHSO_{4}\)) is approximately 1.

pH is a measure of the acidity or basicity of a solution, with a pH of 7 considered neutral, a pH less than 7 considered acidic, and a pH greater than 7 considered basic. The pH is determined by the concentration of hydrogen ions (H+) in the solution.

Sodium bisulfate (\(NAHSO_{4}\)) is a strong acid, meaning that it dissociates completely in water to form hydrogen ions (H+) and sulfate ions (\(SO_{4} ^{2-}\)). The concentration of hydrogen ions in a 0.5 M solution of sodium bisulfate is equal to the concentration of the bisulfate ions, meaning that the pH of the solution would be around 1.

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Newton's third law states that for every action, there is an equal and opposite reaction. This law is applicable in many situations.

The situation where a full cart is harder to move than an empty one is an example of Newton's third law. When a force is applied to the full cart, the cart pushes back with an equal and opposite force making it harder to move.

When a runner has the wind at her back, it is not an example of Newton's third law. This is because the wind is an external force, and there is no direct action-reaction pair involved.

When walking, the foot pushes the ground, and the ground pushes the foot with an equal and opposite force, making it an example of Newton's third law.

When an astronaut moves backward by throwing a hammer forward, it is an example of Newton's third law. When the astronaut throws the hammer forward, the hammer exerts an equal and opposite force on the astronaut, propelling the astronaut backward.

In summary, situations where two objects interact with each other, exerting equal and opposite forces, are examples of Newton's third law.

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The radius of this nucleus is \(4.3054\times 10^{-15}\)m.

A core of the atom that is positively charged. It is made up of two different subatomic particle kinds that are closely clustered. Protons, which have a positive electric charge, and neutrons, which have no electric charge, are the particles.

The nucleus has a radius of about \(10^{-15}\)m. The nucleus has a diameter of about  \(10^{-15}\)m, which is 100,000 times smaller than the diameter of an atom, which is on the order of \(10^{-10}\)m.

The average radius of the nucleus is given by the equation,

\(R = R_{0} A^ {\frac{1}{3} }\)

Here R is the average nuclear radius

\(R_{0}\) is the Fermis constant = \(1.1 \times 10^{15}\)m

A is the mass number = 28 protons + 32 neutrons = 60

Calculating the nuclear radius from mass number:

\(R = 1.1\times 10^{-15} m\times 60^{1/3} \\\)

\(R=1.1\times10^{-15} \times 3.914\)

\(R=4.3054\times 10^{-15}\)m

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As the concentration of NH3 increases, the reaction will shift in the backward direction, causing the concentration of N2 to increase. The correct option is D) The concentration of N2 will increase.

Given the reaction:\(N2(g) + 3H2(g) ⇌ 2NH3(g)\)

The equilibrium constant expression (Kc) for this reaction is:

Kc = \([NH3]2 / [N2] x [H2]3\)

Given Kc = 1.2

Initial concentrations:

[H2]0 = 0.76 M

[N2]0 = 0.60 M

[NH3]0 = 0.48 M

Since there are no NH3 molecules initially, the reaction will proceed in the forward direction. Let x represent the equilibrium concentration of NH3. Thus, the equilibrium concentrations can be expressed as:

\([N2] = [N2]0 - x\)

\([H2] = [H2]0 - 3x\)

\([NH3] = [NH3]0 + 2x\)

Substituting these equilibrium concentrations into the expression for Kc, we get:

Kc = \(([NH3]0 + 2x)2 / ([N2]0 - x) x ([H2]0 - 3x)3\)

Solving this equation gives x = 0.15 M

Therefore, the concentration of NH3 at equilibrium is

0.48 + 2(0.15) = 0.78 M

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ΔH for this reaction : -2668.4 (exothermic)

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

you can search the value of ΔHf on the internet

∆H ° rxn =( ∆H Al₂O₃+∆H AlCl₃+3.∆H NO + 6.∆H H₂O)-(∆H Al+∆H NH₄ClO₄)

\(\tt \Delta H^o~rxn=(-1669.8+-704+(3\times 90.4)+(6.\times -241.8)-(3\times 0+3\times -295)\\\\\Delta H6o~rxn=-3553.4+885=-2668.4~kJ/mol\)

Explanation:

Evolution is a biological process. It is how living things change over time and how new species develop.

Answer: It’s B

Explanation:

Answer:

SOrry it makes me writer sotufjksankdn

Explanation:

Approximately 777.6 grams of water is needed to make 7.2 moles of glucose based on the balanced equation.

The balanced equation provided is:

6CO2 + 6H2O -> C6H12O6 + 6O2

From the equation, we can see that for every 6 moles of water (H2O), 1 mole of glucose (C6H12O6) is produced. Therefore, we need to determine the amount of water required to produce 7.2 moles of glucose.

The mole ratio between water and glucose is 6:1. This means that for every 6 moles of water, we obtain 1 mole of glucose. To find the amount of water needed for 7.2 moles of glucose, we set up a proportion using the mole ratio:

(6 moles H2O / 1 mole glucose) = (x moles H2O / 7.2 moles glucose)

Solving for x, we can cross-multiply:

6 moles H2O * 7.2 moles glucose = x moles H2O * 1 mole glucose

43.2 moles H2O = x moles H2O

Therefore, we need 43.2 moles of water to produce 7.2 moles of glucose.

To convert moles of water to grams, we need to know the molar mass of water, which is approximately 18 g/mol. Using the molar mass, we can calculate the mass of water needed:

Mass of water = moles of water * molar mass of water

Mass of water = 43.2 moles * 18 g/mol

Mass of water = 777.6 g

Therefore, approximately 777.6 grams of water is needed to make 7.2 moles of glucose based on the balanced equation.

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Heat transfer occurs from hot to cold until both objects reach the same temperature. Thus, option 1 is correct.

The second law of thermodynamics states that heat transfer always occurs from hotter objects to cooler objects until they reach the same temperature. This phenomenon is called the Thermal equilibrium of that substance. At this point, both bodies' temperatures are the same.

Heat transfer occurs in three modes. They are conduction, convection, and radiation. In every case, heat is always transferred from got object to sold object. The second law also states that entropy change cannot be negative until they reach equilibrium.

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Answer:

B,C

Explanation:

#4, answer: a This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles.

#5 a, answer: c

The molar absorptivity (ε) of the compound is 2386.36 L mol-1 cm-1.

Molar absorptivity is a measure of how well a substance absorbs light at a specific wavelength. It is represented by the symbol ε (epsilon) and has units of L mol-1 cm-1. The molar absorptivity is defined as the absorbance of a 1 molar solution with a path length of 1 cm. It is also referred to as the molar extinction coefficient or simply the extinction coefficient. The formula for molar absorptivity is as follows:

e = A/(c × l)

where e is the molar absorptivity, A is the absorbance, c is the concentration in molarity, and l is the path length in centimeters.

Given that a 4.4 X 10-5 molar solution has an absorbance of 0.105 when the path length is 1.0 cm, we can calculate the molar absorptivity as follows:

e = A/(c × l) = 0.105/(4.4 X 10-5 × 1.0) = 2386.36 L mol-1 cm-1

Therefore, the molar absorptivity (ε) of the compound is 2386.36 L mol-1 cm-1.

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Answer:

Niels Bohr

Explanation:

Niels Bohr presented an atom model in which the electron could only occupy specific orbits around the nucleus. The electrons were constrained to particular orbits around the nucleus in this atomic model, which was the first to employ quantum theory.

~

Answer:

neils bohr

Explanation:

on in the pic please mark me as brainlist

Explanation:

Potassium is a chemical element with the symbol K and atomic number 19. Potassium is a silvery-white metal that is soft enough to be cut with a knife with little force. Potassium metal reacts rapidly with atmospheric oxygen to form flaky white potassium peroxide in only seconds of exposure

Answer:

Explanation:

The ratio in whole numbers of the masses of carbon that combine with 1.00 g of oxygen between the two compounds is 3:1.

To determine the ratio in whole numbers of the masses of carbon that combine with 1.00 g of oxygen, we need to look at the ratios of their atomic masses.

The atomic mass of carbon (C) is approximately 12.01 g/mol, and the atomic mass of oxygen (O) is approximately 16.00 g/mol.

To find the ratio of their masses, we can divide the atomic masses:

Mass ratio = Atomic mass of carbon / Atomic mass of oxygen

Mass ratio = 12.01 g/mol / 16.00 g/mol

Mass ratio = 0.75125

The ratio obtained, 0.75125, is not a whole number. However, we can approximate it to a whole number ratio by multiplying it by a common factor to eliminate the decimal.

In this case, multiplying the ratio by 4 yields:

Approximate whole number ratio = 4 * 0.75125 ≈ 3

Therefore, the approximate whole number ratio of the masses of carbon that combine with 1.00 g of oxygen is 3:1.

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Answer:

0.46 J/g degrees C, Iron

Explanation:

16.1 J = 5 g x c x 7 degrees C

c = 16.1 / 5 x 7 = 0.46

Answer: 0.46 J/g degrees C, Iron

Answer:

4H₂O(l) → 4H₂(g) + 2O₂(g)

Balancing equations

1. 3Fe(s) + 4H₂O(g) → Fe₃0₄(s) + 4H₂(g)

2. 2AlBr₃(aq) + 3Cl₂(g) → 2AlCl₃(aq) + 3Br₂(l)

3. 2HNO₃(aq) + Ba(OH)₂(aq) → Ba(NO₃)₂(aq) + 2H₂O(l)

4. 2Al(s) + 3Pb(NO₃)₂ → 2Al(NO₃)₃(aq) + 3Pb(s)

5. 3NaOH(aq) + Fe(NO₃)₃ → Fe(OH)₃(s) + 3NaNO₃(aq)

Explanation:

From the question given, I have been able to balance the chemical equations correctly.

In balancing chemical equations, the chemical equation must have equal number of atoms for each element that are both in the reactant side and in the product side.

For the above to be achieved, the number of atoms in each element will have to be multiplied and added.

A look at the balanced chemical equation above, you will discover that each element has the same number of atoms both in the reactant side and in the product side.

Answer:  Mass is 2,37 kg

Explanation: Weight G = mg, and g = 9.81 m/s² on Earth.

m = W/g = 23.2 N /  9.81 m/s²

Answer:

380 mL

Explanation:

A chemist must prepare 925 mL of 325 mM aqueous copper(II) sulfate working solution. He'll do this by pouring out some 0.792 M aqueous copper(II) sulfate stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the copper(II) sulfate stock solution that the chemist should pour out. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Convert "C₂" to M

We will use the conversion factor 1 M = 1000 mM.

325 mM × (1 M/1000 mM) = 0.325 M

Step 3: Calculate the inital volume

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

V₁ = 0.325 M × 925 mL/0.792 M

V₁ = 380 mL

Answer:

option d

Explanation:

isotope means atoms with different mass no. 17 and 18 written on the left represent mass number

this is because atoms can differ in mass no.

In order to find a name of a Hydrocarbon molecule, we need to look for the main carbon chain first, this will be the chain with the highest number of carbon atoms, and in this molecule, we have 4 carbon atoms, therefore we have a 4 carbon molecule, the prefix for this type of molecule is But.

We also have a triple bond in the first carbon, the suffix for triple bonds is "yne"

Therefore, if we add all these informations, we will have 1-butyne, letter A

1. To prepare a 50 ml, 5X Tris glycine buffer from a 100X Tris glycine buffer solution:

Given:

Volume of final buffer (Vf) = 50 ml

Desired concentration of buffer (Cb) = 5X

Concentration of stock buffer (Cs) = 100X

The dilution formula for preparing a desired concentration of a solution is as follows:

Vf * Cb = Vi * Cs

Substituting the given values:

50 ml * 5X = Vi * 100X

Vi = (50 ml * 5X) / 100X

Vi = 2.5 ml

Therefore, you need to take 2.5 ml of the 100X Tris glycine buffer solution and add it to a total volume of 50 ml to obtain a 5X Tris glycine buffer solution.

a. To prepare a 2M solution of NaOH in 100 ml:

Given:

Volume of final solution (Vf) = 100 ml

Desired concentration of NaOH (Cb) = 2M

The molar mass of NaOH is 22.99 g/mol for sodium (Na), 15.999 g/mol for oxygen (O), and 1.007 g/mol for hydrogen (H). Adding these values, we get:

Molar mass of NaOH = 22.99 + 15.999 + 1.007 = 39.996 g/mol ≈ 40 g/mol

2.a.  To calculate the mass of NaOH required for a 2M solution:

Mass (m) = Volume (Vf) * Concentration (Cb) * Molar mass (M)

m = 100 ml * 2M * 40 g/mol

m = 8000 g = 8 kg

Therefore, you need to weigh 8 kg of NaOH and dissolve it in a total volume of 100 ml to obtain a 2M solution of NaOH.

b. To prepare a 10mM solution of NaOH from a 2M solution in 100 ml:

Given:

Volume of final solution (Vf) = 100 ml

Desired concentration of NaOH (Cb) = 10mM = 0.01M

Concentration of stock NaOH solution (Cs) = 2M

Using the dilution formula mentioned earlier:

Vf * Cb = Vi * Cs

Substituting the given values:

100 ml * 0.01M = Vi * 2M

Vi = (100 ml * 0.01M) / 2M

Vi = 0.5 ml

Therefore, you need to take 0.5 ml of the 2M NaOH solution and add it to a total volume of 100 ml to obtain a 10mM NaOH solution.

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The dosage of 26mg can be prepared using approximately 1.625 mL of the 40mg in 2.5 mL solution.

The dosage of 22mg can be prepared using approximately 1.222 mL of the 36mg per 2 mL strength solution.

To calculate the dosage using ratio and proportion, we can set up a proportion based on the strength of the solution.

40mg in 2.5 mL solution will be used to prepare a 26mg dosage.

Let X represent the mL of the solution needed to prepare the 26mg dosage.

We can set up the proportion as follows:

40mg/2.5mL = 26mg/X mL

Cross-multiplying and solving for X, we have:

40mg * X mL = 2.5mL * 26mg

40X = 65

X = 65/40

X ≈ 1.625 mL

For the second question:

36mg per 2 mL strength solution is used to prepare 22mg.

Let Y represent the mL of the solution needed to prepare the 22mg dosage.

We can set up the proportion as follows:

36mg/2mL = 22mg/Y mL

Cross-multiplying and solving for Y, we have:

36mg * Y mL = 2mL * 22mg

36Y = 44

Y = 44/36

Y ≈ 1.222 mL

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Answer:

An example of a chemical change is wood burning.

Explanation:

A chemical change is something that emits light, puts off heat, emits a scent, changes the chemical properties of an item, or some/all of the above.

Chemical changes occur when a substance combines with another to form a new substance, called chemical change.

A chemical change occurs when one substance is transformed into one or more new products via a chemical reaction.

1. burning of paper.

2. cooking of food

3. burning of wood

4. ripening of fruits

5. rotting of fruits.

6. frying egg

7. rusting of iron

8. mixing acid and base.

9. burning of the candle

10. leaves changing colour

11. melting of sugar.

12. baking

13. the explosion of fireworks.

14. souring milk

15. digestion of food

16.fermentation

17. lighting matchstick

18. photosynthesis

19. decomposition of waste

20. making popcorn

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