2al + 6 hcl → 2 alcl3 + 3h2 ∆hrxn = -152 kj

how much heat energy is associated with the reaction of 35 g of aluminum with excess hydrochloric acid?

Answer:

I don’t know what recording station you’re referring to but, When seismic waves reach the seismograph, a graphical record, or seismogram, is produced

Explanation:

The seismic waves hitting the recording station has been resulted in the seismograph, that has been evident of the earthquake.

The seismic wave has been the radiation, with the result of the movement of the earth surface. The movement has been result in the earthquake.

The intensity of the earthquake has been measured by seismograph on the Richter scale. The seismic wave results in the movement of the leads to the production of the seismograph.

The seismic waves hitting the recording station has been resulted in the seismograph, that has been evident of the earthquake.

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9) The reaction is spontaneous as written

10) The reaction is not spontaneous as written

A spontaneous reaction is a chemical reaction that occurs without the need for an external energy input, such as heat or electricity. It is a reaction that occurs naturally, without any outside assistance, and proceeds on its own without the need for additional energy to be added to the system.

It is important to note that a spontaneous reaction does not necessarily mean that the reaction will occur quickly or with a high rate. The rate of a reaction depends on various factors, such as the concentration of reactants, temperature, and the presence of catalysts.

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Answer:

A part.

Explanation:

Because the reactants must be the exact same as the the products.

Answer:

i. The face-centered cubic (FCC) crystal structure consists of a unit cell that contains 4 atoms.

ii. Copper exhibits the FCC crystal structure, and the atomic packing factor (APF) for copper can be determined.

i. In a face-centered cubic (FCC) structure, each corner of the unit cell is occupied by one atom, and each face of the unit cell is occupied by half an atom. Therefore, there are a total of 8 corners, with each contributing 1/8th of an atom, and 6 faces, with each contributing 1/2 an atom. The total number of atoms per unit cell in an FCC structure is calculated as (8 x 1/8) + (6 x 1/2) = 4 atoms.

ii. The atomic packing factor (APF) is a measure of how efficiently the atoms are packed in a crystal structure and is calculated as the ratio of the volume occupied by the atoms to the total volume of the unit cell. For an FCC crystal structure, the APF can be determined using the formula:

APF = (Number of atoms per unit cell) x (Volume of each atom) / (Volume of the unit cell)

To determine the APF of copper, specific values for the number of atoms per unit cell and the volume of the unit cell and atom are required. Unfortunately, the specific values for copper are not provided in the question.

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From the calculation, the binding energy of the oxygen atom is 1.98 * 10^11 J

The term binding energy has to do with the energy that must be supplied to the nucleus of an atom for the nucleus of the atoms to be bonded together.

We must note that;

E = mc^2

E =  2.20 × 10 -28 kg * ( 3 * 10^8m/s)^2 = 1.98 * 10^11 J

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Answer:

1.98 × 10 -11 J

Explanation:

edg

Answer:Proton, the gold foil experiment indicated that the nucleus has a positive charge.

The substance that contains a hydroxide group and dissociates to produce hydroxide ions in aqueous solutions is called a base.

A base is a type of chemical substance that has a pH greater than 7 and reacts with an acid to form a salt and water. The hydroxide group, which is also known as the hydroxyl group, is a functional group that contains an oxygen atom and a hydrogen atom.

This group is often written as OH, and it is responsible for the basic properties of a substance. When a substance that contains a hydroxide group is dissolved in water, it dissociates to produce hydroxide ions, which have a negative charge and a formula of OH-.

Examples of common bases include sodium hydroxide (NaOH), potassium hydroxide (KOH), and calcium hydroxide (Ca(OH)2). These substances are often used in chemical reactions, as well as in the production of soaps, detergents, and other products.

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Answer:

D

Explanation:

I just took the quiz on a pex

The electrode is the anode in a galvanic cell should be option d. The electrode with the highest oxidation potential.

The electrode should be lies in the left half-cell to represent the anode since oxidation should be arise here.

Here the flow of anions should be in the salt that bridges towards it. In the right half-cell that represent the cathode since the reduction arise here.

Hence, the option d is correct.

And the rest of the options are wrong.

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To calculate the maximum mass of zinc that will react with 50 cm3 of hydrochloric acid with a concentration of 2.0 mold/dm3, we first need to use the balanced chemical equation for the reaction between zinc and hydrochloric acid:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
From this equation, we can see that one mole of zinc reacts with 2 moles of hydrochloric acid. Therefore, the number of moles of hydrochloric acid in 50 cm3 (0.05 dm3) with a concentration of 2.0 mold/dm3 is:
n(HCl) = C × V = 2.0 × 0.05 = 0.1 mol
Since the stoichiometry of the reaction is 1:2 (1 mole of zinc reacts with 2 moles of hydrochloric acid), the maximum number of moles of zinc that can react is:
n(Zn) = 0.1/2 = 0.05 mol
Finally, we can use the molar mass of zinc (65.38 g/mol) to calculate the maximum mass of zinc that can react:
mass(Zn) = n(Zn) × M(Zn) = 0.05 × 65.38 = 3.27 g
Therefore, the maximum mass of zinc that will react with 50 cm3 of 2.0 mold/dm3 hydrochloric acid is 3.27 grams.

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The amine that has the given spectroscopic data is N,N-dimethylisopropylamine. To identify the amine, we will analyze the provided 1H NMR and 13C NMR data.

1H NMR:
- 1.15 ppm (9H singlet): This indicates three methyl groups (\(CH_{3}\)) present in the molecule, each contributing three protons.
- 1.41 ppm (2H broad singlet): This indicates a methylene group (\(CH_{2}\)) attached to a nitrogen atom (amine group).
13C NMR:
- 32.6 ppm: This corresponds to a carbon that is attached to the nitrogen atom (amine) and is part of the methylene group (\(CH_{2}-N\)).
- 47.4 ppm: This corresponds to the carbon that has three methyl groups (\(CH_{3}\)) attached to it.
Taking these pieces of information together, we can deduce the structure of the amine: N,N-dimethylisopropylamine. This molecule has one \(CH_{2}-N\) group, with the nitrogen atom bonded to two methyl groups and a central carbon atom connected to three methyl groups.
Based on the provided 1H NMR and 13C NMR spectroscopic data, the amine in question is N,N-dimethylisopropylamine.

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The stable isotopes have an infinite half-life. It's hard to get longer than infinite. There are 252 stable isotopes.

The longest known half-life is currently tellurium-128 at 7.7 × 10^24 years.

The longest directly measured half-life is presently xenon-124 with 1.8 × 10^22 years.

It is somehow possible that some isotopes that we currently think are stable might turn out to have a very long half-life, but we haven't figured it out yet.

For example, scientists used to think that bismuth-209 was stable.

In 2003 we discovered it is radioactive with a half-life of 2.01 × 10^19 years.

Stable isotopes are “stable" and therefore not radioactive. An atomic nucleus needs to be “unstable” for it to be radioactive.

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Answer:

D

Explanation:

The first ionization energy is the minimum energy required to remove the most loosely bound electron from an atom.

Good luck!

Answer:

Come all bad girls

Come all girls to see and show

fhg-hxcr-ujn

Answer:

See Explanation

Explanation:

Let the mass of salt be x

Let the mass of solution be 10 g

Percentage by mass of salt = 0,25%

So;

Percent concentration of salt = mass of salt/mass of solution * 100

0.25 = x/10 * 100

0.25 * 10/100 = x

x= 0.025 g of salt

Let the mass of water be y

Since mass of solution = mass of salt + mass of water

10 g = 0.025 g + y

y = 10 g - 0.025 g

y = 9.975 g

Answer:

400x

Explanation:

magnification=magnification of objective lens*magnification of eyepiece lens

m=40*10

m=400x

Answer:

The particles in a gas are moving very quickly in random directions.

The particles in a solid are tightly packed and locked in place, and they are vibrating in their place.

The particles in a liquid are close together (touching) but they are able to move/slide/flow past each other.

brainliet please?

x

Answer:

Calcium and magnesium sulphide react to form calcium sulphide and magnesium metal.

Explanation:

Ca + MgS → CaS + Mg

The reaction between calcium and magnesium sulfide produces calcium sulfide and magnesium metal. The balanced equation can be given as Ca + MgS → CaS + Mg.

An equation for just a chemical reaction is said to be balanced if both the reactants as well as the products have the same number of atoms and total charge for each component of the reaction. In other words, both side of the reaction have an equal balance of mass and charge.

Conservation of charge as well as mass, equation and reaction balance, etc. The components and outcomes of a chemical reaction are listed in an imbalanced chemical equation, but the amounts necessary to meet the conservation of mass are not specified. The reaction between calcium and magnesium sulfide produces calcium sulfide and magnesium metal. The balanced equation can be given as Ca + MgS → CaS + Mg.

Therefore, the balanced equation can be given as Ca + MgS → CaS + Mg.

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In order to calculate the boiling point of benzene (C6H6) using thermodynamic information, we need to understand the concept of boiling point.

In order to calculate the boiling point of benzene (C6H6) using thermodynamic information, we need to understand the concept of boiling point. Boiling point is the temperature at which a substance changes from a liquid to a gas state. It is determined by the intermolecular forces between the molecules of the substance.
The ALEKS data tab provides thermodynamic information such as the enthalpy of vaporization (ΔHvap) and the boiling point of the substance at standard pressure (1 atm). For benzene, the ΔHvap is 30.8 kJ/mol and the boiling point at 1 atm is 80.1 °C.
Using the Clausius-Clapeyron equation, we can relate the boiling point of a substance to its enthalpy of vaporization and its vapor pressure. However, since we do not have the vapor pressure of benzene, we cannot use this equation directly.
Instead, we can use the fact that the boiling point of a substance is directly proportional to the vapor pressure of the substance. This means that if we know the boiling point at one pressure, we can use the Antoine equation to calculate the boiling point at a different pressure.
For benzene, we can use the Antoine equation:
log10(P) = A - (B / (T + C))
where P is the vapor pressure in mmHg, T is the temperature in Kelvin, and A, B, and C are constants.
We can rearrange this equation to solve for the temperature (T) at a given vapor pressure (P). For standard pressure (760 mmHg), the boiling point of benzene is 80.1 °C. Using this value and the Antoine constants for benzene (A = 6.90565, B = 1211.033, and C = 220.79), we can solve for the boiling point at a different pressure.
For example, if we want to know the boiling point of benzene at 500 mmHg, we can plug in P = 500 and solve for T:
log10(500) = 6.90565 - (1211.033 / (T + 220.79))
T = 344.9 K = 71.7 °C
Therefore, the boiling point of benzene at 500 mmHg is approximately 72 °C (rounded to the nearest degree).

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(a) The rate law for the decomposition of N₂O₅ is rate = k[N₂O₅]², and the value of k is 1.60 × 10⁻³ s⁻¹.

(b) The concentration of N₂O₅(g) at 250 s is 0.0180 M. The calculated answer makes sense as it falls within the range of concentrations observed during the reaction.

(c) The concentration of N₂O₅(g) at 600 s is 0.00338 M. The calculated answer makes sense as it shows a decrease in concentration over time.

(d) The concentration of N₂O₅(g) is equal to 0.00150 M at approximately 377 s. This is determined by setting up an equation using the rate law and solving for time.

(a) To determine the rate law, we analyze the given data. By comparing the initial rate with different concentrations of N₂O₅, we find that the rate is proportional to [N₂O₅]². Therefore, the rate law is rate = k[N₂O₅]². To calculate the value of k, we can use any set of data points and substitute them into the rate equation along with the corresponding time values. By rearranging the equation to solve for k, we can determine its value to be 1.60 × 10⁻³ s⁻¹.

(b) To find the concentration of N₂O₅(g) at 250 s, we can interpolate between the given data points of time and concentration. By considering the rate equation, we can calculate the concentration of N₂O₅(g) at 250 s to be 0.0180 M. This calculated answer makes sense because it falls within the range of concentrations observed during the reaction.

(c) Similar to part (b), we can interpolate between the data points to find the concentration of N₂O₅(g) at 600 s. By applying the rate equation and calculations, the concentration is determined to be 0.00338 M. This answer makes sense as it indicates a decrease in concentration over time due to the decomposition reaction.

(d) To find the time at which the concentration of N₂O₅(g) is equal to 0.00150 M, we can rearrange the rate equation to solve for time. By substituting the given concentration value into the equation and solving for time, we find that the concentration is equal to 0.00150 M at approximately 377 s.

This can be verified by substituting the time value back into the rate equation and confirming that the calculated concentration matches the given value.

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The correct answer is: c. 5.7 mg. It's important to note that when performing calculations involving proportions, it's crucial to use consistent units.

To determine the amount of dye needed when using only 65 mL of water, we can set up a proportion based on the given information.

Amount of dye = 22 mg

Amount of water = 250 mL

New amount of water = 65 mL

Unknown amount of dye = ?

We can set up the proportion as follows:

(22 mg) / (250 mL) = (x mg) / (65 mL)

To solve for x (the unknown amount of dye), we can cross-multiply and solve for x:

22 mg * 65 mL = 250 mL * x mg

1430 mg·mL = 250x mg·mL

To isolate x, we divide both sides of the equation by 250 mL:

(1430 mg·mL) / (250 mL) = (250x mg·mL) / (250 mL)

5.72 mg = x

Therefore, when using only 65 mL of water, we will need 5.72 mg of dye.

In this case, both the amount of dye and the amount of water were given in milligrams (mg) and milliliters (mL), respectively. By setting up the proportion correctly and performing the calculation, we determined the required amount of dye for the given volume of water.

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Explanation:

moles = mass/mr

=60.3/28

=2.15 moles

number of particles = moles × 6.02×10^23

= 2.15×6.02×10^23

= 1.29×10^24 atoms

CO=12+16=28

Answer:

2.29 × 10²⁴ atoms Na

Explanation:

Step 1: Define

Avagadro's Number: 6.02 × 10²³ atoms, molecules, formula units, etc.

Step 2: Stoichiometry

\(3.80 \hspace{3} mol \hspace{3} Na(\frac{6.02(10)^{23} \hspace{3} atoms \hspace{3} Na}{1 \hspace{3} mol \hspace{3} Na} )\) = 2.2876 × 10²⁴ atoms Na

Step 3: Simplify

We have 3 sig figs.

2.2876 × 10²⁴ atoms Na ≈ 2.29 × 10²⁴ atoms Na

Answer:

The charge of a Strontium atom that has 10 electrons is 4.48 × 10⁻¹⁸ C

Explanation:

The atomic number of Strontium is 38

The electronic configuration of Strontium is given as follows;

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²

The atomic number of Strontium is 38 electrons in orbit and 38 protons in its nucleus and is electronically neutral

38e⁻ + 38e⁺ = 0

When there are only 10 electrons in an atom of Strontium, we have;

10e⁻ + 38e⁺ = +28e

The charge of a Strontium atom that has 10 electrons is +28e

+28e = 28×1.6×10⁻¹⁹ C = 4.48 × 10⁻¹⁸ C

The charge of a Strontium atom that has 10 electrons is +28e or 4.48 × 10⁻¹⁸ C.

the basic unit of structure and function in all living things.

Answer:

all living things.

Explanation:

The rate of appearance of I3- under the given experimental conditions is 37.4 x 10-3 M/s, which is the same as the rate of disappearance of I-.

The rate of appearance of I3- can be determined using the stoichiometry of the reaction. From the balanced equation, we can see that 1 mole of I3- is formed for every 1 mole of I- that disappears. Therefore, the rate of appearance of I3- will be the same as the rate of disappearance of I-.

Given that the initial rate of disappearance of I- is 37.4 x 10-3 M/s, we can conclude that the rate of appearance of I3- under the same experimental conditions is also 37.4 x 10-3 M/s.

In conclusion, the rate of appearance of I3- under the given experimental conditions is 37.4 x 10-3 M/s, which is the same as the rate of disappearance of I-.

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