Characteristics that are used to help identify an object are called

Answer:

The structures in the cell work together to perform its life functions

Explanation:

Answer:

What statement best explains how life functions a unicellular organism are carried out?​

Explanation:

Answer:  [tex][H^+][/tex] of 0.056 M HF solution is [tex]8.96\times 10^{-5}[/tex]

Explanation:

[tex]HF\rightarrow H^+F^-[/tex]

 cM              0             0

[tex]c-c\alpha[/tex]        [tex]c\alpha[/tex]          [tex]c\alpha[/tex]  

So dissociation constant will be:

[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Give c= 0.056 M and [tex]\alpha[/tex] = ?

[tex]K_a=1.45\times 10^{-7}[/tex]

Putting in the values we get:

[tex]1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}[/tex]

[tex](\alpha)=0.0016[/tex]

[tex][H^+]=c\times \alpha[/tex]

[tex][H^+]=0.056\times 0.0016=8.96\times 10^{-5}[/tex]  

Thus [tex][H^+][/tex] of 0.056 M HF solution is [tex]8.96\times 10^{-5}[/tex]

Answer:

Am I supposed to answer these on brainly?

Explanation:

Answer:

You can better relate to the air temperature.

Explanation:

Fahrenheit gives you almost double the precision of Celsius without having to use decimals.

The  ion(s) would NOT be present in the net ionic equation will be[tex]Pb^{2+[/tex], [tex]NO^{3-}[/tex]

Option (a) is correct.

To determine which ions would not be present in the net ionic equation, we need to identify the spectator ions. Spectator ions are the ions that do not participate in the chemical reaction and remain unchanged throughout the reaction.

The net ionic equation represents the chemical equation after removing the spectator ions. We can determine the spectator ions by comparing the initial and final compounds and identifying which ions remain the same on both sides of the reaction.

Let's analyze the given reaction:

[tex]Pb(NO_3)_2(aq) + 2KI(aq)[/tex] → [tex]PbI_2(s) + 2KNO_3(aq)[/tex]

The balanced equation shows that [tex]Pb^{2+[/tex] and [tex]2NO_3^-[/tex] ions combine with 2K+ and 2I- ions to form [tex]PbI_2(s)[/tex] and [tex]2KNO_3(aq)[/tex]. In the reaction, the [tex]NO_3^-[/tex]ions are part of both the starting compound [tex](Pb(NO_3)_2)[/tex]and the product compound [tex](KNO_3)[/tex]. Therefore, the [tex]NO_3^-[/tex] ions are spectator ions and would not be present in the net ionic equation.

Now, let's consider the answer choices:

A) [tex]Pb^{2+}, NO_3[/tex]-: [tex]NO_3^-[/tex] ions are spectator ions, so this option is correct.

B) [tex]K+, NO_3[/tex]-:[tex]NO_3^-[/tex] ions are spectator ions, so this option is correct.

C) [tex]K^+, Pb^{2+}[/tex]: Both [tex]K^+[/tex] and [tex]Pb^{2+}[/tex] ions participate in the reaction, so this option is incorrect.

D) [tex]K^+, I^-[/tex]: Both [tex]K^+[/tex] and[tex]I^-[/tex] ions participate in the reaction, so this option is incorrect.

Therefore, the correct answer is:

A) [tex]Pb^{2+}, NO_3^-[/tex]

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Answer:

Question 4 is- Solubility

Question 5 is- Suspension

Hopes this helps >:D

Answer:

10.Semi-Solid

11.Liquid

12.Solid

13:Semi-Solid

Explanation:

Answer:

See explanation

Explanation:

Oxidation half equation;

3Cd(s) + 6OH^-(aq) ------> 3Cd(OH)2(s) + 6e

Reduction half equation;

2MnO4^-(aq) + 8H^-(aq) + 6e --------> 2MnO2(s) + 4H2O(l)

Balanced reaction equation;

3Cd(s) + 6OH^-(aq) + 2MnO4^-(aq) + 8H^-(aq)  ------> 3Cd(OH)2(s) + 2MnO2(s)

Number of electrons transferred = 6

If the Co (NO₃)₂ solution contains about 6 g Co₂+ per liter, then there are 60 micrograms of Co₂+ ions in one spot.

To calculate the number of micrograms of Co₂+ ions in one spot, we need to convert the given concentration from grams per liter (g/L) to micrograms per microliter (µg/µL) and then multiply it by the volume of the spot.

Given:

Volume of solution for one spot = 10 µL

Concentration of Co₂+ ions in the solution = 6 g/L

First, we need to convert the concentration from grams per liter to micrograms per microliter:

6 g/L = 6 × (1E+6) µg/L (since 1 g = 1E+6 µg)

= 6 × (1E+6) µg / (1E+6) µL (since 1 L = 1E+6 µL)

= 6 µg/µL

Now, we can calculate the number of micrograms of Co₂+ ions in one spot:

Number of micrograms of Co₂+ ions = Concentration of Co₂+ ions × Volume of solution for one spot

= 6 µg/µL × 10 µL

= 60 µg

Therefore, there are 60 micrograms of Co₂+ ions in one spot.

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Answer:

neutral

Explanation:

There are three basic subatomic particles. These are;

A neutron has no charge unlike the proton and the electron. It is present in the nucleus and contributes to the mass of the atom.

The comparison between an electrical circuit and a water circuit can be helpful in understanding the concepts and principles of electricity by drawing parallels with a familiar system like the flow of water.

In both circuits, the potential energy or pressure that drives the flow is represented by voltage or PSI. Just as pipes provide a path for water to flow, conductors in an electrical circuit provide a path for electricity. The pump in a water circuit acts as the source of energy, similar to a battery in an electrical circuit. Both valves and switches control or regulate the flow by either opening or closing the circuit or pathway. Restrictions in a water circuit and resistance in an electrical circuit impede the flow and reduce the overall current or flow rate. The water meter and ammeter measure the flow rate or current passing through the circuit. Water itself in a water circuit and electrons in an electrical circuit act as carriers of energy. The high-pressure output and positive voltage represent the part of the circuit with higher potential energy, while the low-pressure intake and negative voltage represent the part with lower potential energy. When a valve is closed, it corresponds to an open circuit, interrupting the flow or current. Conversely, when a valve is open, it can be compared to a closed circuit, allowing the flow or current to pass through. The flow rate in a water circuit, measured in liters/second, is similar to the current in an electrical circuit, measured in amps.

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Answer:86.84

Explanation:

The standard entropy change for the given reaction is -1052.0 J/K. The entropy change of the reaction is negative, which indicates that the reaction is not spontaneous.

Given, Reaction: 4 NH3 (g) + 5 O2 (g) à 4 NO (g) + 6 H2O (g)SubstanceS° (J/mol × K)NH3 (g)192.8O2 (g)205.2NO (g)210.8H2O (g)188.8The formula for finding the standard entropy of reaction is as follows:ΔS°rxn = ΣS°products – ΣS°reactantsWhere, ΔS°rxn = Standard entropy changeS°products = Sum of the standard entropies of the productsS°reactants = Sum of the standard entropies of the reactants. For the given reaction, the standard entropy change is calculated by finding the difference between the sum of standard entropies of products and the sum of standard entropies of reactants.ΔS°rxn = ΣS°products – ΣS°reactants= [4(S°NO) + 6(S°H2O)] – [4(S°NH3) + 5(S°O2)]= [4(210.8) + 6(188.8)] – [4(192.8) + 5(205.2)]= 843.6 – 1895.6= -1052.0 J/K. Thus, the standard entropy change for the given reaction is -1052.0 J/K. The entropy change of the reaction is negative, which indicates that the reaction is not spontaneous.

In the given question, we are asked to calculate the standard entropy change, ΔS°rxn. The formula for finding the standard entropy of reaction is:ΔS°rxn = ΣS°products – ΣS°reactantsWhere, ΔS°rxn = Standard entropy changeS°products = Sum of the standard entropies of the productsS°reactants = Sum of the standard entropies of the reactants. By using this formula and substituting the values from the table, we get:ΔS°rxn = ΣS°products – ΣS°reactants= [4(S°NO) + 6(S°H2O)] – [4(S°NH3) + 5(S°O2)]= [4(210.8) + 6(188.8)] – [4(192.8) + 5(205.2)]= 843.6 – 1895.6= -1052.0 J/K. Thus, the standard entropy change for the given reaction is -1052.0 J/K.

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Answer:

32÷5

I'm just tryna get points I'm sorry

goodluck tho❤

In a dissolving metal reduction, ammonia serves as a solvent, and each sodium atom acts as an electron donor. As a result, the alkyne is reduced to give an alkene product.

Dissolving metal reduction is a technique used to reduce alkynes to alkenes using an alkali metal, such as lithium or sodium, in liquid ammonia. During this process, the alkali metal dissolves in liquid ammonia to produce a deep blue color, which is a result of electrons being released from the alkali metal.

This process results in the formation of an intermediate solution containing free electrons that act as strong reducing agents, which can reduce the alkyne to an alkene. Therefore, in this reaction, ammonia serves as a solvent, while each sodium atom acts as an electron donor, leading to the reduction of the alkyne to form an alkene product.

Finally, we can say that in a dissolving metal reduction, ammonia serves as a solvent, and each sodium atom acts as an electron donor. As a result, the alkyne is reduced to give an alkene product.

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The idea of Continental drift explains the presence of 300-million-old glacial grooves on four separate southern continents as they were once joined together and subject to the same climate conditions.

The theory of Continental drift states that the continents were once joined together as a supercontinent called Pangaea and have since drifted apart. This movement has caused the formation of geological features and altered the climate of the earth. 300-million-old glacial grooves have been found on four separate southern continents. This suggests that the continents were once joined together and were subject to the same climate conditions at the time of the formation of these grooves. These continents could have been connected through land bridges or narrow passages, which allowed for the migration of flora and fauna. The movement of these landmasses could have been caused by tectonic activity, which can be linked to the theory of Continental drift.

In conclusion, the idea of Continental drift explains the presence of 300-million-old glacial grooves on four separate southern continents as they were once joined together and subject to the same climate conditions.

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I'm hoping the same for my new coaching classes

This is my first time going out to study ngl-

Answer:

please give me brainlist and follow

Explanation:

Contamination of ground water can result in poor drinking water quality, loss of water supply, degraded surface water systems, high cleanup costs, high costs for alternative water supplies, and/or potential health problems. The consequences of contaminated ground water or degraded surface water are often serious.

The Gibbs Free Energy at 298 K for the reaction of nitrogen and hydrogen to form ammonia is -149.2128 kJ/mol. Since this value is negative, it indicates that the reaction is spontaneous under standard conditions (i.e., at 298 K and 1 atm pressure)

The Gibbs Free Energy is one of the most important thermodynamic functions used to determine whether a chemical reaction is spontaneous or not at a given temperature and pressure. It is represented by the symbol "ΔG" and is defined as the difference between the enthalpy (ΔH) and entropy (ΔS) of a system at a constant temperature and pressure. The formula for Gibbs Free Energy is: ΔG = ΔH - TΔS, where T is the temperature in Kelvin (K).

The reaction of nitrogen and hydrogen to form ammonia can be represented by the following chemical equation: N2(g) + 3H2(g) → 2NH3(g)

To calculate the Gibbs Free Energy at 298 K for this reaction, we need to know the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) for the reaction. These values can be found in a standard thermodynamic data table or by using Hess's Law to calculate them from known enthalpies of formation.

For this reaction, the standard enthalpy change is -92.2 kJ/mol and the standard entropy change is +191.6 J/mol-K. Therefore, we can calculate the Gibbs Free Energy at 298 K using the formula:

ΔG° = ΔH° - TΔS°

= (-92.2 kJ/mol) - (298 K)(0.1916 kJ/mol-K)

= -92.2 kJ/mol - 57.0128 kJ/mol

= -149.2128 kJ/mol

Thus, the Gibbs Free Energy at 298 K for the reaction of nitrogen and hydrogen to form ammonia is -149.2128 kJ/mol. Since this value is negative, it indicates that the reaction is spontaneous under standard conditions (i.e., at 298 K and 1 atm pressure).

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Answer:

N₂O

Explanation:

Empirical formula is defined as the simplest whole number ratio of atoms presents in a molecule

For a nitrogen oxide that contains 63.66% of nitrogen, the percent of oxygen must be:

100-63.66 = 36.34% Oxygen

This percent is the percent in mass. To solve this question we must convert the mass of each atom to moles in order to find the simplest whole number ratio:

Moles of the atoms:

N = 63.66g * (1mol / 14g) = 4.547 moles N

O = 36.34g * (1mol / 16g) = 2.271 moles O

The ratio N:O is:

4.547 moles N / 2.271 moles O = 2

That means there are 2 atoms of N per atom of O and the empirical formula is:

The term "atomic theory" is used instead of "atomic hypothesis" because it signifies the evolution and acceptance of the concept over time.

While Democritus initially proposed the idea of atoms in 400 BCE, it was merely a hypothesis without substantial experimental evidence. Over centuries, scientific investigations and advancements led to a deeper understanding of atomic structure and behavior.

The term "atomic theory" acknowledges that the concept of atoms has undergone refinement and modification based on experimental evidence and theoretical developments.

It recognizes that the understanding of atoms has evolved from a speculative hypothesis to a well-established scientific theory supported by extensive experimental observations, mathematical models, and empirical data.

The term "theory" conveys the comprehensive and validated nature of our understanding of atoms, encompassing their complex structures and behavior.

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To determine the value of ΔSo (change in entropy) for the given reaction, we need to consider the difference in the number of moles of gas between the reactants and the products.

Reactants:

2 moles of C2H2 (g)

5 moles of O2 (g)

Products:

4 moles of CO2 (g)

2 moles of H2O (l)

The change in the number of moles of gas is given by Δn = (moles of gas in products) - (moles of gas in reactants).

Δn = (4 moles of CO2 + 2 moles of H2O) - (2 moles of C2H2 + 5 moles of O2)

= 4 - 2 + 2 - 5

= -1

The ΔSo value can be calculated using the equation ΔSo = ΣnΔSo(products) - ΣnΔSo(reactants).

Since Δn is -1, we have:

ΔSo = (4 mol x ΔSo(CO2) + 2 mol x ΔSo(H2O)) - (2 mol x ΔSo(C2H2) + 5 mol x ΔSo(O2))

Assuming the standard entropy values, we have:

ΔSo = (4 mol x 213.7 J/(mol·K) + 2 mol x 69.9 J/(mol·K)) - (2 mol x 200.8 J/(mol·K) + 5 mol x 205.0 J/(mol·K))

= 122.3 J/K

Therefore, the value of ΔSo for the given reaction is +122.3 J/K.

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The combustion of 9.8 moles of methanol will produce 352.8 grams of water.

First, let us look at the balanced equation of combustion of methanol:

[tex]2CH_3OH(l)+3O_2(g)-- > 2CO_2(g)+4H_2O(l)[/tex]

The ratio of number of moles of methanol burned and that of water produced is 1:2.

Thus, 9.8 moles of methanol will produce 19.6 moles of water.

Mass of 19.6 moles water = 19.6 x 18 = 352.8 grams

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The IUPAC names of the following organic compounds are correct with the given naming conventions:A. Pent-3-eneB. Prop-1-en-2-yneC. 1-methylpropaneD. All are incorrect - This is the incorrect option because all the given options have correct IUPAC names of the organic compounds. Hence, option D is incorrect.

The International Union of Pure and Applied Chemistry (IUPAC) is an organization that establishes a set of rules for the naming of chemical compounds. This is done to make sure that all scientists in the world use the same names for the same compounds. Therefore, the names should be unique and unambiguous. The IUPAC name of a compound provides information about its molecular structure, functional groups, and substituents. Some of the examples are given below:A. Pent-3-ene - It is a five-carbon molecule with a double bond between the third and fourth carbons. Hence, the name of the compound is pent-3-ene.B. Prop-1-en-2-yne - It is a three-carbon molecule with a triple bond between the first and second carbons and a double bond between the second and third carbons. Hence, the name of the compound is prop-1-en-2-yne.C. 1-methylpropane - It is a three-carbon molecule with one methyl group attached to the first carbon. Hence, the name of the compound is 1-methylpropane.

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Answer:

c

Explanation:

1. litmus paper is used when showing a change between a greater range in ph levels - so A and D are automatically a no.

2. methyl red is used to show a range in ph levels between 4.8-6

3. Option C is the only one left so im going to assume its C because its definitely not A, B, or D

Answer:

b because  i said so

Explanation:

In one of the trials, the purpose of using water/soap solution was to compare the cleanliness of the hand with washing by water alone.

Hand washing is one of the simplest, most effective ways to avoid getting sick and prevent the spread of germs. Washing your hands with soap and water is still one of the most important steps you can take to avoid getting sick and to avoid spreading germs to others. The purpose of using water/soap solution for one of the trials was to compare the cleanliness of the hand with washing by water alone.The experiment involves two trials to investigate the effectiveness of soap and water in removing bacteria from hands. In one trial, the participant washed their hands with soap and water. While in the other trial, the participant washed their hands with water alone. After washing, their hands were pressed on a petri dish with culture medium to grow the bacteria. Then, the plates were placed in an incubator at 37°C for two days to grow bacteria. The soap and water solution are effective in removing bacteria from hands because the soap helps to lift dirt, grease, and microbes off skin and onto the surfaces of the lather, so that it can be rinsed away by water.

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The mass of barium in 288 mg of barium arsenate (Ba3(AsO4)2) is approximately 0.462 mg.

The molar mass of Ba3(AsO4)2 can be calculated by summing the atomic masses of each element in the compound. The atomic masses are as follows:

Barium (Ba): 137.33 g/mol

Arsenic (As): 74.92 g/mol

Oxygen (O): 16.00 g/mol

Molar mass of Ba3(AsO4)2:

= (3 * 137.33 g/mol) + (2 * (74.92 g/mol + 4 * 16.00 g/mol))

= 411.99 g/mol + 207.84 g/mol

= 619.83 g/mol

Now, we can calculate the mass of barium in 288 mg of barium arsenate using the molar mass and the given mass.

Mass of barium:

= (mass of barium arsenate / molar mass of barium arsenate) * molar mass of barium

= (288 mg / 619.83 g/mol) * 411.99 g/mol

= 0.462 mg

The mass of barium in 288 mg of barium arsenate (Ba3(AsO4)2) is approximately 0.462 mg.

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if the halo of our galaxy is spherically symmetric and has a constant mass density, the mass density remains the same at all radii within the halo. This assumption simplifies the calculation, allowing us to consider a uniform mass density throughout the spherically symmetric halo.

If the halo of our galaxy is spherically symmetric, we can make certain assumptions about its mass distribution. Let's consider a simplified model where the halo has a constant mass density throughout its volume.

In this case, the mass density (ρ) represents the amount of mass per unit volume. To calculate the mass density, we divide the total mass of the halo by its volume. However, since we do not have the specific values for the total mass or the volume, we will express the mass density in terms of an equation.

Let's denote the total mass of the halo as M and the volume of the halo as V. The mass density (ρ) is then given by:

ρ = M / V

Since we assume the halo is spherically symmetric, we can use the formula for the volume of a sphere:

V = (4/3)πr³

where r represents the radius of the sphere.

To obtain the mass density as a function of the radius (ρ(r)), we need to find an expression for the total mass (M) in terms of the radius.

Assuming a constant mass density throughout the halo, the mass (M) enclosed within a sphere of radius r is given by:

M = ρ * V = ρ * (4/3)πr³

Substituting this expression for M into the equation for mass density, we have:

ρ(r) = (ρ * (4/3)πr³) / ((4/3)πr³)

Simplifying the equation, we find that the mass density is constant and independent of the radius:

ρ(r) = ρ

Therefore, if the halo of our galaxy is spherically symmetric and has a constant mass density, the mass density remains the same at all radii within the halo. This assumption simplifies the calculation, allowing us to consider a uniform mass density throughout the spherically symmetric halo.

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