Help. Why are state symbols used in chemical equations?
O A. They tell which reactions will happen and which won't.
B. They identify how much product will be made.
C. They identify what phase the substances are in.
O D. They tell how the atoms are arranged in the substances.

The concentration of 39% could not be obtained when mixing a 35% acid solution with a 45% acid solution.

When mixing two acid solutions, the resulting concentration will fall between the concentrations of the initial solutions. In this case, the chemist is mixing a 35% acid solution and a 45% acid solution. The resulting concentration will range between 35% and 45%. Therefore, concentrations of 37% and 43% can be obtained.

However, a concentration of 39% cannot be achieved since there is no intermediate concentration between 35% and 45% which equals 39%. The resulting concentration will always be a weighted average of the two initial concentrations. Additionally, a concentration of 49% cannot be obtained as it exceeds the highest initial concentration (45%).

The resulting concentration cannot exceed the concentration of the most concentrated solution used. Hence, the concentration that cannot be obtained is 39%, while options 37%, 43%, and 49% are achievable concentrations by mixing the 35% and 45% acid solutions.

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Answer:

zany

Explanation:

Answer:

a fresh start

Explanation:

a. When the temperature is decreased, the reaction will shift to the right in order to produce more heat and counteract the decrease in temperature.
b. When the volume is decreased, the reaction will shift to the left in order to decrease the pressure.
4. To calculate the pH of a 0.25 M solution of propylamine (C3H7NH2),

we need to use the equation:
Kb = [C3H7NH3+][OH-]/[C3H7NH2]
6.9*10-4 = (x)(x)/(0.25)
x = 0.0132 M
pOH = -log[OH-] = -log(0.0132) = 1.88
pH = 14 - pOH = 14 - 1.88 = 12.12
Therefore, the pH of the solution is 12.12.
5. To calculate the pH of a 0.20 M solution of Ba(CN)2, we need to use the equation:
Ka = [HCN][OH-]/[CN-]
4.4*10-5 = (x)(x)/(0.20)
x = 0.0021 M
pH = -log[H+] = -log(0.0021) = 2.68
Therefore, the pH of the solution is 2.68.

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Plate tectonic processes are responsible for various crustal features, such as mountains at convergent boundaries, mid-ocean ridges at divergent boundaries, earthquakes and volcanic activity at transform boundaries, and visible features like rift valleys and subduction zones.

The three types of plate boundaries are convergent, divergent, and transform boundaries.

Many crustal features can be seen as having resulted from plate tectonics. Where two plates come together at a convergent boundary, they collide and form such features as mountains. When two plates move away from each other the result is a divergent boundary, and they form such features as mid-ocean ridges. A transform boundary forms when two plates slide past each other. If the pressure between two plates builds high enough, the plates may move rapidly causing earthquakes and the occurrence of volcanic activity. Features such as ridges and trenches can be seen from space and lend support to the plate tectonics theory.

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If a process is irreversible, it means that the entropy of the system increases.

The concept of irreversibility is related to the directionality of a process. A reversible process is one that can be reversed without any net increase in entropy. However, an irreversible process is one that cannot be reversed without an increase in the entropy of the system and its surroundings. In other words, during an irreversible process, the entropy of the system increases, and energy is dissipated into the surroundings, resulting in a loss of usable energy. Therefore, option A, "the entropy of the system increases" is the correct answer.

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Answer:

I believe it is 9. mass divided by volume and to find volume multiply LxWxH

Answer:

Ok

Explanation:

When compound a is treated with bromine in the presence of uv light, the major product is 2,2-dibromopentane is 2S-2bromo pentane.

Nucleophile, A reactant that offers a pair of electrons to form a brand new covalent bond (i.e. a Lewis base). A nucleophile that shares an electron pair with a proton is usually referred to as a Bronsted-Lowry base, or only a base.

NaSH  gives SH-  nucleophile which attacks in SN2 manner  i.e from opposite plane . If product has R configuration then reactant should have S confiuration.

We have Br attached to 2nd carbon  since  treatment with Br2/hv  gave 2,2 dibromo compound.

Hence we had  2S-2bromo pentane.

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The molecular shapes which have the most electron pairs (both shared and unshared) around the central atom are linear-shaped.

The molecular shapes are the shapes like linear, trigon planner, octahedral, tetrahedral, etc. which depend on the number of atoms and the lone pair and bond pair of atoms.

In linear shape, a maximum number of electrons take part in the hybridization and can also have lone pairs to give repulsion to the bond pairs of the atoms.

Therefore,  linear-shaped are the molecular shapes that have the most electron pairs (both shared and unshared) around the central atom.

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Propene on reaction with N-bromosuccinimide in CCl4 produces 3-bromopropene.

In this reaction, allylic H atom is replaced with Br atom.

An allylic radical is the kind of reactive intermediate that is created when propene reacts with n-bromosuccinimide (nbs) to produce 3-bromo-1-propene. With NBS as one of the reactants, it is likely that a free radical bromination will take place.

The methyl group in propene would be attacked because the free radical that is forming can only be stabilized via resonance. A radical is said to be allylic if its resonance forms, which individually include unpaired electrons, are all located on an allylic carbon. Depending on where the allylic carbon is located, it can be categorized as a primary, secondary, or tertiary allylic radical.

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The specific heat capacity of the nickel metal is 0.45.

Let us recall that the specific heat capacity is used to describe the amount of the quantity of heat that is able to cause the temperature of a unit mass of a substance to increase by 1 kelvin.

Now we know that the heat that is lost by the metal is equal to the heat that is gained by the water as such we have;

mcwdT = -mcmdT

cm = mcwdT/mdT

cm = Heat capacity of the metal

cw = Heat capacity of the water

m = mass of the metal and the water

dT = temperature change

To obtain the heat capacity;

cm = 150 * 4.2 * (25 - 23.5)/-28.2 * (25.0 - 99.8)

cm = 945/2109.4

cm = 0.45

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Answer:

The basic unit of any element would be the atom.

Answer:

A.

Explanation:

Particles in gas move very fast and quickly while particles in liquids are slow and closer together.

The total volume of the solution is 1005 mL after adding 5 mL of 0.50 M HCl, we can assume that the volume change is negligible, and the concentrations of HA and A- remain the same.

A buffer typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer's ability to resist changes in pH comes from the equilibrium between the weak acid and its conjugate base.

To calculate the pH after adding 5.0 mL of 0.50 M HCl to 1 liter (1000 mL) of the buffer solution, we need to consider the buffer's composition and the effect of the added acid.

Since you haven't provided the exact composition of the buffer, I'll assume it consists of a weak acid and its conjugate base. Let's denote the weak acid as HA and its conjugate base as A-. The buffer is designed to resist changes in pH when small amounts of acid or base are added.

When HCl is added, it will dissociate into H+ and Cl- ions. The H+ ions from HCl will react with the conjugate base A- in the buffer, forming the weak acid HA. This reaction helps maintain the pH of the buffer.

To calculate the pH after the addition of HCl, we need to know the initial concentrations of HA and A- in the buffer and the pKa of the weak acid HA. With this information, we can use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

Given that the total volume of the solution is 1005 mL after adding 5 mL of 0.50 M HCl, we can assume that the volume change is negligible, and the concentrations of HA and A- remain the same.

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After adding 5.0 mL of 0.50 M HCl to 1 liter of a buffer solution with an initial concentration of 0.10 M and a \(pKa\) of 5.0, the pH of the buffer will be determined using the Henderson-Hasselbalch equation: \(pH = 5.0 + log([0.10 M + (0.50 M x (5.0 mL/1005 mL))]/[0.10 M])\).

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which is \(pH = pKa + log([A-]/[HA])\).
In this case, 5.0 mL of 0.50 M HCl is added to 1 liter of the buffer solution, resulting in a total volume of 1005 mL. To calculate the pH, we need to know the \(pKa\) value of the buffer and the concentrations of the acidic form \(([HA])\) and the conjugate base \(([A-])\).
Let's assume the \(pKa\) of the buffer is 5.0. We can use the equation to calculate the ratio \([A-]/[HA]\). Since the concentration of HCl added is much higher than the initial concentration of the buffer, we can assume that the concentration of [HA] remains approximately the same. Therefore, the concentration of [A-] will be equal to the initial concentration of the buffer plus the concentration of HCl added.
Let's say the initial concentration of the buffer is 0.10 M. After adding 5.0 mL of 0.50 M HCl, the concentration of [A-] would be 0.10 M + (0.50 M x (5.0 mL/1005 mL)).
Now, substitute the values into the Henderson-Hasselbalch equation: \(pH = 5.0 + log(([A-]/[HA]))\). Calculate the ratio \([A-]/[HA]\) and substitute it into the equation to find the \(pH\).
This calculation will give you the pH of the buffer solution after adding the HCl.

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Answer:

2.5

Explanation:

13.4 by 5.36

Mass by volume

Answer:

11.1 grams per cubic cm

Explanation:

Volume = 13.4 in^3 = 219.587 cm^3

Mass = 5.36 lb = 2431.255 gram

\(density = \frac{mass}{volume} \\ = \frac{2431.255}{219.587} \\ = 11.0719441 \\ \approx \: 11.1 \: grams \: per \: {cm}^{3} \)

The 250 liters of gas that is collected in the expandable, sealed container. The sample is then heated from the 15.0 °C to the 45.0 °C at the constant pressure. The new volume of the container is 226.4 L.

The temperature and the volume at constant pressure is as :

V₁ / T₁ = V₂ / T₂

V₂ = V₁ T₂ / T₁

The initial volume of the gas, V₁ = 250 L

The final volume of the gas, V₂ = ?

The initial temperature of the gas, T₁ = 15 + 273

The initial temperature of the gas, T₁ = 288 K

The final temperature of the gas, T₂ = 45 + 273

The final temperature of the gas, T₂ = 318

V₂ = V₁ T₂ / T₁

V₂= ( 250 × 288 ) / 318

V₂ = 226.4 L

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Hydroxide ions are produced in solution by Arrhenius bases. There are no free hydrogen ions in water. Because bases in water form hydroxide ions, it will result in their production.

Ka=[H+][A−]/[HA] .The amount of the original acid that has been ionised in solution is represented by the acid ionisation. As a result, the numerical value of Ka represents the acid's strength. Stronger acids than weaker acids with relatively lower Ka values include weak acids with substantially higher Ka values.

When an acid is dissolved in water, hydrogen ions are produced; as a result, the solution's hydrogen ion concentration rises.

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Answer:

Where are the answer choices

?

First nations people along the west coast of British Columbia used smelting technology to create copper plaques is false.

Smelting process is the form of extractive metallurgy to produce a metal from its ore. it is the process of melting and separation of charges .  it is extensive energy process. Plaques are made by pouring molten metal in to a mold. and when it gets cools , it solidified into a copper solid plaque. first nation people use copper as hunting weapons, spear tools etc. copper is metal.

Thus, First nations people along the west coast of British Columbia used smelting technology to create copper plaques is false.

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decay rate of approximately 1.69x10^17 Bq (becquerels),

The decay rate of a radioactive sample is determined by the number  of radioactive atoms present and the decay constant, which represents the probability of decay per unit of time.

To calculate the decay rate, we multiply the number of atoms in the sample by the decay constant. In this case, the sample has 8.31x10^17 atoms and a decay constant of 0.203/s. Multiplying these values gives a decay rate of approximately 1.69x10^17 Bq (becquerels), which represents the number of decays per second in the sample.

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Potassium Metal Is Explosive— Do Not Use It! The reaction of sodium with water is a spectacular and essential classroom demonstration. Many teachers want to show also the more violent reaction of potassium. We propose not to do so because explosions can happen even before the metal is in contact with water.

- BRAINLIEST answerer

Answer:

Explanation:

Potassium and cold water are not used to prepare hydrogen in laboratory because: 1. Potassium is highly reactive metal and it reacts violently with cold water. 2.

Answer:

thermal shock

Explanation:

the temperatures inside the glass jar should have continued to increase over time. Internal stresses due to uneven heating. This is also known as “thermal shock”.

In general, the thicker the glass, the more prone it will be to breaking due to the immediate differences in temperature across the thickness of glass.

Borosilicate glass is more tolerant of this, as it has a higher elasticity than standard silicon glass.

You may also note that laboratory test tubes and flasks are made with thinner walls, and of borosilicate glass, when designated for heating.

Answer:

each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element.

Explanation:

Answer: Weather currents change wind tempature and there for change the weather of the seasons.

Answer:

The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases equals the original mass of the charcoal and the oxygen when it first reacted.

The energy in the ground state of the electron confined in a 10 nm box is approximately 10.89 eV, and the energy in the first excited state is approximately 43.56 eV.

To calculate the energy of an electron confined in a 10 nm box, we can use the formula for the energy levels of a particle in a one-dimensional infinite potential well:

E_n = (n^2 * h^2) / (8 * m * L^2)

where:

E_n is the energy of the nth energy level,

n is the quantum number of the energy level (n = 1 for the ground state),

h is the Planck's constant (6.626 x 10^-34 J·s),

m is the mass of the electron (9.10938356 x 10^-31 kg),

L is the length of the box (10 nm = 10 x 10^-9 m).

Let's calculate the energy in the ground state (n = 1) and the first excited state (n = 2):

For the ground state (n = 1):

E_1 = (1^2 * h^2) / (8 * m * L^2)

Substituting the values:

E_1 = (1^2 * (6.626 x 10^-34 J·s)^2) / (8 * (9.10938356 x 10^-31 kg) * (10 x 10^-9 m)^2)

Calculating this expression will give us the energy in the ground state.

For the first excited state (n = 2):

E_2 = (2^2 * h^2) / (8 * m * L^2)

Substituting the values:

E_2 = (2^2 * (6.626 x 10^-34 J·s)^2) / (8 * (9.10938356 x 10^-31 kg) * (10 x 10^-9 m)^2)

Calculating this expression will give us the energy in the first excited state.

Please note that the energies calculated will be in joules (J). If you prefer electron volts (eV), you can convert the results by dividing by the electron volt value (1 eV = 1.602 x 10^-19 J).

Performing the calculations:

For the ground state:

E_1 = (1^2 * (6.626 x 10^-34 J·s)^2) / (8 * (9.10938356 x 10^-31 kg) * (10 x 10^-9 m)^2) ≈ 1.747 x 10^-18 J

For the first excited state:

E_2 = (2^2 * (6.626 x 10^-34 J·s)^2) / (8 * (9.10938356 x 10^-31 kg) * (10 x 10^-9 m)^2) ≈ 6.987 x 10^-18 J

Converting the energies to electron volts (eV):

E_1 ≈ 10.89 eV (rounded to two decimal places)

E_2 ≈ 43.56 eV (rounded to two decimal places)

Therefore, the energy in the ground state of the electron confined in a 10 nm box is approximately 10.89 eV, and the energy in the first excited state is approximately 43.56 eV.

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