4. Which of the following is NOT true? Select all that apply
a. Gases are made of individual particles; either atoms or molecules
b. Particles are close together even when they do not collide
c. Energy is not transferred to the walls of the container when particles collide with the container walls
d. Gas in is constant motion

most bacteria have flagellum, also nerve cells are larger

Rounding these values to the nearest whole number, we get the empirical formula of ibuprofen as C6H9O.

To determine the empirical formula of ibuprofen, we need to convert the mass percent composition into mole ratios. This can be done by assuming that we have 100 grams of ibuprofen, and calculating the number of moles of each element present in that sample.

Starting with carbon, we have 75.69 grams of carbon in our sample, which corresponds to 6.30 moles (using the atomic weight of carbon). Similarly, we have 8.80 grams of hydrogen, which corresponds to 8.74 moles, and 15.51 grams of oxygen, which corresponds to 0.97 moles.

To get the simplest whole number ratio of these elements, we divide each mole value by the smallest one (0.97):

- Carbon: 6.30 / 0.97 = 6.49
- Hydrogen: 8.74 / 0.97 = 9.00
- Oxygen: 0.97 / 0.97 = 1.00

This means that the molecular formula of ibuprofen could be a multiple of this empirical formula (e.g. C12H18O2), but we would need additional information (such as the molecular weight) to determine that.

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A chemical labeled as bacteriostatic will inhibit bacterial growth but will not kill them.

A bacteriostatic chemical will slow down or inhibit the growth and reproduction of bacteria but will not kill them. This is different from bactericidal chemicals which are capable of killing bacteria. Bacteriostatic chemicals are often used in medical settings to control the growth of bacteria while the body's immune system fights off the infection. It is important to note that the effectiveness of bacteriostatic chemicals varies depending on the type of bacteria and the concentration of the chemical used.

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When the impetus driving an object decreases, its motion also decreases. And when the impetus is completely removed, the object stops moving.

When the impetus driving an object decreases, its motion also decreases. The term "impetus" in this context refers to the force that sets an object in motion or maintains its motion. When this force decreases, the object experiences a decrease in its velocity or acceleration. This is due to the fact that the force acting on the object is directly proportional to the rate of change of its motion, as described by Newton's second law of motion.

If the impetus is completely removed, the object stops moving altogether. This is because there is no longer any force acting on the object to maintain its motion, and hence it decelerates and eventually comes to rest. This can be seen in everyday scenarios, such as a ball rolling to a stop when it reaches the bottom of a hill or a car slowing down and stopping when the engine is turned off.

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--The complete question is, What happens to the motion of an object when the impetus driving it decreases, and what happens when the impetus is completely removed?--

The half-life of the reaction at 36°C is 19.2 seconds.

We can use the following equation to determine the  half-life of the reaction at 36°C;

[tex]t_{1/2}[/tex] = ln(2) / k

where [tex]t_{1/2}[/tex] is the half-life of the reaction and k is the reaction rate constant at the given temperature.

First, we need to find the reaction rate constant at 36°C. We can use the two rate constants given for 30°C and 42°C and the Arrhenius equation;

ln(k₂/k₁) = (-Ea/R) × (1/T₂ - 1/T₁)

where k₁ and k₂ are the rate constants at temperatures T₁ and T₂, Ea will be the activation energy, R is gas constant, and T is temperature in Kelvin.

We can choose 30°C (303 K) as T₁ and 42°C (315 K) as T₂, and solve for ln(k₂/k₁) to get;

ln(k₂/k₁) = (-Ea/R) × (1/T₂ - 1/T₁)

ln(0.0608/0.0202) = (-Ea/8.314 J/(mol×K)) × (1/315 K - 1/303 K)

Ea ≈ 52.7 kJ/mol

Next, we can use the Arrhenius equation to find the rate constant at 36°C (309 K);

k = A × exp(-Ea/RT)

k = 0.0202 s⁻¹ × exp(-52.7 kJ/mol / (8.314 J/(mol×K) × 309 K))

k ≈ 0.036 s⁻¹

Finally, we can use the half-life equation with this rate constant to find the half-life at 36°C;

[tex]t_{1/2}[/tex]= ln(2) / k

[tex]t_{1/2}[/tex] = ln(2) / 0.036 s⁻¹

[tex]t_{1/2}[/tex] ≈ 19.2 s

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Coinage metals, which typically include copper, silver, and gold, are chosen over more active metals for use in coins because they are less reactive and more resistant to corrosion.

This ensures durability and preserves the appearance of the coins. Some more active metals like zinc or nickel are sometimes used in coins due to their lower cost and availability, while still maintaining adequate resistance to corrosion and wear for everyday use.

The reason why the last 4 miles in the activity series of metals, which are gold, silver, platinum, and palladium, are commonly referred to as the "coinage medals" is because they are highly resistant to corrosion and have a low reactivity towards other chemicals, making them ideal for use in coins. These metals are also very rare and valuable, which adds to their appeal as a currency.

More active metals such as zinc or nickel are sometimes used in coins because they are more abundant and less expensive than the "coinage metals". However, these metals tend to be more reactive and therefore more prone to corrosion and other chemical reactions, which can affect the appearance and value of the coins over time. Additionally, the use of these metals in coins is often limited to lower denominations or commemorative coins, rather than as a standard currency.

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The "coinage metals" are typically gold, silver, copper, and platinum, which are the last 4 metals in the activity series. These metals are chosen over more active metals for use in coins because they are relatively unreactive and do not corrode easily, making them ideal for coins that need to be durable and long-lasting. Additionally, these metals have been historically valued and used as currency, making them culturally significant as well.

However, some more active metals such as zinc or nickel are sometimes used in coins because they are cheaper and more readily available than the coinage metals. These metals may be used as an alloy with the coinage metals to make coins more affordable, or they may be used as a substitute for the more expensive metals in lower denomination coins. However, these metals are not as durable as the coinage metals and may corrode more easily, leading to shorter lifespans for the coins.

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The answer is 1.3 x 10^-5 for the Ka of propanoic acid.

To solve this problem, we can use the relationship between the pH, the concentration of the acid, and the acid dissociation constant (Ka).

First, we need to find the concentration of propanoic acid in the solution. We know that 0.100 moles of propanoic acid are dissolved in 1.00 L of solution, so the concentration is:

concentration = 0.100 mol / 1.00 L = 0.100 M

Next, we can use the Ka value to set up an expression for the acid dissociation reaction:

HC3H5O2 + H2O ⇌ C3H5O2- + H3O+

The Ka expression for this reaction is:

Ka = [C3H5O2-][H3O+] / [HC3H5O2]

We can assume that the concentration of C3H5O2- is equal to the concentration of H3O+, since they are produced in a 1:1 ratio. Let's call this concentration x. Then the concentration of HC3H5O2 will be (0.100 - x), since some of the acid has dissociated.

Substituting these values into the Ka expression and solving for x gives:

Ka = x^2 / (0.100 - x) = 1.3 x 10^-5
x = 1.47 x 10^-3 M

Now we can use the definition of pH to find the pH of the solution:

pH = -log[H3O+] = -log(x) = -log(1.47 x 10^-3) = 2.832

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In Tollen's test, the reaction of aldehydes with silver(i) ions in basic solution results in the formation of silver metal and carboxylate.

Specifically, the reaction involves the oxidation of the aldehyde and the reduction of the silver(i) ion. This can be seen in the reaction of 2 silver 1 ions with a generic aldehyde and 3 hydroxide ions, which produces 2 silver atoms, a generic carboxylate, and 2 water molecules. The species being oxidized in the reaction is the aldehyde, while the species being reduced is the silver(i) ion. The visual indicator of a positive test is the formation of silver metal, which indicates the presence of an aldehyde in the sample.

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In this Tollen's test, the species being oxidized is the aldehyde (RCHO), while the species being reduced is the silver(I) ion (Ag+). The visual indicator of a positive test is the formation of silver metal (Ag), which appears as a shiny silver mirror on the inner surface of the test tube.

In the Tollen's test, the reaction involves aldehydes reacting with silver(I) ions in a basic solution to form silver metal and a carboxylate. The generic equation for this reaction is:

2 Ag+ + RCHO + 3 OH- → 2 Ag + RCOO- + 2 H2O

In the Tollen's test, aldehydes react with silver(i) ions in basic solution to form silver metal and a carboxylate. The reaction involves the oxidation of the aldehyde and reduction of the silver(i) ion. Specifically, in the presence of 2 silver(i) ions and 3 hydroxide ions, a generic aldehyde is oxidized to form a generic carboxylate and 2 water molecules, while the silver(i) ions are reduced to form 2 silver atoms. The visual indicator of a positive test is the formation of silver metal, which indicates the presence of an aldehyde. Therefore, in this reaction, the aldehyde species is being oxidized.

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Answer:

At STP 1 mol of any gas occupies a volume of 22.4 liters. So the calculation is:

206.08/22.4 = 9.2 mol

The correct option is

In an endothermic response(reaction), the whole vitality(total energy) at the beginning of the response is more noteworthy than the full vitality at the conclusion of the response 

because endothermic responses retain warmth from the environment, which implies that the vitality of the framework increments.

An endothermic response may be a chemical response that retains warmth from the environment, which implies that the vitality of the framework increments.

This increment in vitality is utilized to break the bonds between the particles or atoms within the reactants, and the items are shaped from the modification of these iotas or atoms into unused bonds.

As a result, the whole vitality of the framework at the conclusion of the response is more noteworthy than the full vitality at the start of the response. This increment in vitality is ordinarily watched as an increment within the temperature of the framework or its environment. 

In an endothermic response, the whole vitality at the beginning of the response is less than the overall vitality at the end of the response

.

Usually, endothermic responses retain warmth from the environment, which implies that the vitality of the framework increments.

As a result, the entire vitality of the framework at the conclusion of the response is greater than the full energy at the start of the response. Subsequently,

The proper reply is "In an endothermic response(reaction), the whole vitality(total energy) at the beginning of the response is more noteworthy than the full vitality at the conclusion of the response ".

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The freezing and boiling points of 0.060 m [tex]MgCl_2[/tex] are -0.33°C and 100.09 °C. 0.060 m  [tex]FeCl_3[/tex] has the following freezing and boiling points of -0.44°C and 100.12 °C respectively.

Depression in the freezing point and elevation in the boiling point are colligative properties. Colligative properties refer to the properties that are dependent on the concentration of solute in the solution.

Depression in the freezing point is calculated as ΔT = [tex]ik_fm[/tex]

where ΔT is depression in the freezing point

i is the dissociation factor

[tex]k_f[/tex]  is the freezing depression factor = 1.86°C kg/mol

m is the molality of the solution

So, depression in 0.060 m [tex]MgCl_2[/tex] is 3*1.86*0.06

( it has 3 as a dissociation factor as it breaks into 1 [tex]Mg^{2+[/tex] and 2 [tex]Cl^-[/tex] ions)

0 - freezing point = 0.33

freezing point = -0.33°C

So, depression in 0.060 m [tex]FeCl_3[/tex] is 4*1.86*0.06

( it has 4 as a dissociation factor as it breaks into 1 [tex]Fe^{3+[/tex] and 3 [tex]Cl^-[/tex] ions)

0 - freezing point = 0.44

freezing point = -0.44°C

Elevation in boiling point is calculated as ΔT = [tex]ik_bm[/tex]

where ΔT is Elevation in boiling point

i is the dissociation factor

[tex]k_b[/tex]  is the boiling elevation factor = 0.51°C kg/mol

m is the molality of the solution

So, elevation in 0.060 m [tex]MgCl_2[/tex] is 3*0.51*0.06

( it has 3 as a dissociation factor as it breaks into 1 [tex]Mg^{2+[/tex] and 2 [tex]Cl^-[/tex] ions)

boiling point - 100 = 0.09

boiling point = 100.09 °C

So, elevation in 0.060 m [tex]FeCl_3[/tex] is 4*0.051*0.06

( it has 4 as a dissociation factor as it breaks into 1 [tex]Fe^{3+[/tex] and 3 [tex]Cl^-[/tex] ions)

boiling point - 100 = 0.12

boiling point = 100.12 °C

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The density of the glass fragment is approximately 2.26 g/ml

To calculate the density of the glass fragment, we can use the formula:

Density = Mass / Volume

First, let's calculate the volume of the glass fragment using the displacement method. The volume of water displaced when the glass fragment was submerged in the graduated cylinder is given as 1.14 ml.

So, the volume of the glass fragment is 1.14 ml.

Next, we can calculate the density of the glass fragment by dividing the mass of the glass fragment by its volume:

Density = Mass / Volume = 2.573 g / 1.14 ml

Density = 2.573 g / 1.14 ml ≈ 2.26 g/ml

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Expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.

Aluminum is chemical element with symbol Al and atomic number is 13.

4Al + 3O₂ → 2Al₂O₃

10 g Al × 1 mol Al / 26.98 g Al = 0.371 mol Al

4 g O₂ × 1 mol O₂ / 32.00 g O₂ = 0.125 mol O₂

We determine the limiting reactant by comparing the mole ratios of aluminum and oxygen in the balanced equation and reactant that produces  smaller amount of product is limiting reactant. In this case, aluminum is the limiting reactant because it produces only 0.1855 moles of aluminum oxide, which is less than the 0.25 moles of aluminum oxide produced by the oxygen:

0.371 mol Al × 2 mol Al₂O₃ / 4 mol Al = 0.1855 mol Al₂O₃

0.125 mol O₂ × 2 mol Al₂O₃ / 3 mol O2 = 0.2083 mol Al₂O₃

0.1855 mol Al₂O₃ × 101.96 g/mol = 18.93 g Al₂O₃

Therefore, expected grams of aluminum oxide product from the given masses of reactants are 18.93 g.

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The  electrons cannot have any arbitrary energy within the atom, and they can be given enough energy to escape the atom, forming ions.

Electrons in an atom change energy by moving between discrete energy levels, which are quantized states within the atom. These energy levels are determined by the electron's orbitals and the principal quantum number.

Electrons can gain or lose energy through processes like absorption or emission of photons, respectively. When an electron gains enough energy, it can jump to a higher energy level, or

even escape the atom, resulting in ionization. Conversely, when an electron loses energy, it transitions to a lower energy level, emitting a photon in the process.

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A chemical reaction has a Q10 of 3 option  c. a rate of 9 at 20°C and 3 at 30°C is the rates that characterizes this reaction

The Q10 value is a measure of how much the rate of a chemical reaction changes with a 10°C change in temperature. A Q10 of 3 indicates that the rate of the reaction will increase by a factor of 3 when the temperature is raised by 10°C.

Looking at the answer choices, we can see that option a and b have a Q10 value of 2, which is not the same as the given Q10 value of 3. Option e has a Q10 value of 4, which is also not the same.

Option d has a Q10 value of 3, but the rates given are at 20°C and 40°C, which is not a 10°C change in temperature.

Therefore, the only option that fits the given Q10 value and has rates that are 10°C apart is option c, which has a rate of 9 at 20°C and 3 at 30°C. Therefore, the answer is c.

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Option c states that the rate of the reaction is 9 at 20°C and 3 at 30°C. The ratio of rates between 20°C and 30°C is 9/3 = 3, which matches the Q10 value of 3.  

c. a rate of 9 at 20°C and 3 at 30°C

The Q10 value is a measure of the temperature sensitivity of a reaction, and it is defined as the factor by which the rate of a reaction changes for every 10-degree Celsius change in temperature. A Q10 value of 3 indicates that the rate of the reaction increases by a factor of 3 for every 10-degree Celsius increase in temperature.

This means that the rate of the chemical  reaction is consistent with the temperature sensitivity indicated by the given Q10 value, making option c the correct answer.

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There are 0.00799 moles of gold present in the sample of crushed rock.

The formula to convert the number of atoms of an element to moles is:

moles = number of atoms / Avogadro's number

where Avogadro's number is approximately 6.022 x 10^23.

Using the given information, we can calculate the number of moles of gold present in the sample:

moles of gold = 4.81 x 10^21 atoms / 6.022 x 10^23 atoms/mol

moles of gold = 0.00799 mol

Note: The answer has been rounded to five significant digits in accordance with the significant figures of the given number of atoms.

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One of the most popular names for this grape variety is "Mission grape".

Yes, a grape variety introduced in Chile by European Catholic missionaries is known by a variety of names. One of the most popular names for this grape variety is "Mission grape", which is believed to have originated from the Catholic missionaries who brought the grape to Chile. However, the grape variety is also known by other names such as Pais, Criolla Chica, Listan Prieto, and many others depending on the region and the local dialects. Despite the different names, this grape variety remains an important part of Chile's viticulture history and is still widely cultivated in the country today.

A grape variety introduced in Chile by European Catholic missionaries is known as the "País" grape, also referred to as "Mission" grape or "Criolla Chica." This grape variety was brought to Chile by Catholic missionaries in the 16th century to produce wine for religious ceremonies.

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The grape variety introduced in Chile by European Catholic missionaries is known by a variety of names, but it is most commonly referred to as the "Mission grape."

European Catholic refers to the Catholic Church in Europe, which has a long and complex history. The Catholic Church was the dominant religious institution in Europe during the medieval period, with its influence extending into political and cultural spheres. Throughout the centuries, the Church played a significant role in shaping the continent's religious, social, and political landscape.

The Church's teachings, doctrines, and traditions were transmitted through the continent's various societies and cultures, and many of Europe's greatest art, music, and architecture have been inspired by the Catholic faith. The Catholic Church has also been involved in significant political events in European history, such as the Crusades, the Reformation, and the Counter-Reformation. Today, the Catholic Church remains a significant presence in Europe, with over 200 million Catholics living on the continent. It continues to be an essential institution in shaping European culture and values.

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Option (C) is correct. One should count the atoms of each element on both sides of the chemical equation to make sure they are equal and decide whether the equation is balanced and appropriately constructed.

The number of moles of a substance created or consumed during the chemical reaction is indicated by the coefficients next to the entity symbols.

Verify each answer to verify sure the result does not result in the original equation's denominator being equal to zero. a denominator in the original equation can be made zero if a solution can be discovered.

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Question:

What is a method you could use to determine if an equation is written in the correct way and balanced?

One method to determine if an equation is written in the correct way and balanced is to check that the number and type of atoms are the same on both sides of the equation by using the Law of Conservation of Mass.

Scientific notation is a way of expressing very large or very small numbers using powers of 10.

The general form of a number in scientific notation is:

a × 10^b

where;

To convert a number to scientific notation, follow these steps:

Identify the decimal point in the number. If the number is an integer, assume the decimal point is at the end of the number (e.g., 100 is the same as 100.0).

Move the decimal point to the right or left so that only one non-zero digit remains on the left side of the decimal point.

Count the number of places the decimal point moved. This will be the value of "b" in the scientific notation.

The remaining number on the left side of the decimal point is "a" in the scientific notation.

Write the number in the form "a × 10^b".

Here's an example:

Number: 2450

Identify the decimal point: 2450.

Move the decimal point to the left after the first digit: 2.450.

Count the number of places the decimal point moved (in this case, 3 places to the left): b = 3.

The remaining number on the left side of the decimal point is 2.45: a = 2.45.

Write the number in scientific notation: 2.45 × 10^3.

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The molar mass of the gas is approximately 65.3 g/mol. The closest answer choice is 65.3 g/mol, so that is the correct answer.

To calculate the molar mass of the gas, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to calculate the number of moles of the gas using the given information:

n = (PV) / (RT)

n = (1.04 atm * 0.854 L) / (0.0821 L·atm/(mol·K) * 302 K)

n = 0.0361 mol

Next, we can calculate the molar mass of the gas by dividing its mass by the number of moles:

molar mass = mass / number of moles

molar mass = 2.34 g / 0.0361 mol

molar mass = 64.9 g/mol

Therefore, the molar mass of the gas is approximately 65.3 g/mol. The closest answer choice is 65.3 g/mol, so that is the correct answer.

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The volume of the gas at 175.0 °C will be 24.50 Litres

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

V₁/T₁ = V₂/T₂

Where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

To use this formula, we need to convert the temperatures to Kelvin by adding 273.15 to them:

T1 = 20°C + 273.15 = 293.15 K

T2 = 175.0°C + 273.15 = 448.15 K

Substituting the values into the formula, we get:

16.00 L / 293.15 K = V2 / 448.15 K

Solving for V2, we get:

V2 = 24.50 L

Therefore, the final volume is 24.50 L.

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The grams of N₂ would have to react to produce 31.5 kJ of the energy is 84 g.

The chemical equation is as :

N₂ + 3H₂   --->     2NH₃       ΔH = -96 kJ

The energy produces = 31.5 kJ

We have multiplied the factor so that the value of the enthalpy has also been multiplied.

The factor = 96 / 31.5 = 3

Thus, the balanced chemical equation is :

3N₂ + 9H₂   --->   6NH₃  

The moles of N₂ = 3 mol

The mass of the N₂ = moles × molar mass

The mass of the N₂ = 3 mol × 28 g/mol

The mass of the N₂ = 84 g

The amount of the N₂ would have to react to produce the 31.5 kJ of the energy is 84 g.

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This question is incomplete, the complete question is :

The following thermochemical equation is for the reaction of n2 with h2 to form nh3. how many grams of n2 would have to react to produce 31.5 kj of energy?

N₂ + 3H₂   --->     2NH₃      ΔH = -96 kJ

The residue removal log is used every time you rinse or air dry to remove residue from equipment before using it with organics.

The residue removal log is used whenever unloading a chlorine dishwasher, as it helps track the process of ensuring that equipment is free of residue before using it with organics.

Removal of logging residue negatively affected tree diameter and height, but had no significant effect on the basal area of the subsequent stand (in the mid-term). On the other hand, different methods of mechanical site preparation (bedding, plowing furrows, and trenching) had no effect on tree growth 1 year after planting, but had a significant effect on tree diameter, tree height, and basal area in the mid-term. Bedding treatments could have a significant positive impact on the productivity of the subsequent Scots pine stands, even when planted on sandy, free-draining soils.

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Mike needs to dilute 0.5 liters of the 1.2 M NaOH stock solutions to reach his desired 2-liter, 0.3 M final solution.

To make a 2 liter, of 0.3 M NaOH solution, Mike needs to dilute his 1.2 M stock solution. The calculation can be done using the formula:

C1V1 = C2V2

where C1 is the concentration of the stock solution, V1 is the volume of the stock solution to be diluted, C2 is the desired concentration of the final solution, and V2 is the total volume of the final solution.

Rearranging the formula to solve for V1, we get:

V1 = (C2V2) / C1

Plugging in the values, we get:

V1 = (0.3 M x 2 L) / 1.2 M

V1 = 0.5 L

Therefore, Mike needs to dilute 0.5 liters of his 1.2 M stock solution with water to make a 2 liter, 0.3 M NaOH solution.

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Mike needs to dilute 0.5 liters of his 1.2 M stock solution to make 2 liters of a 0.3 M NaOH solution for his experiment.

To make a 2 liter, 0.3 M NaOH solution, Mike will need to dilute his 1.2 M stock solution.

The equation for dilution is: C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

We can rearrange this equation to solve for V1, which is the volume of stock solution that Mike needs to dilute.

C1V1 = C2V2
V1 = (C2V2)/C1

In this case, C1 = 1.2 M, C2 = 0.3 M, and V2 = 2 liters.

V1 = (0.3 M * 2 liters)/1.2 M = 0.5 liters

So Mike needs to dilute 0.5 liters of his 1.2 M stock solution to make 2 liters of a 0.3 M NaOH solution for his experiment.

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Fill the volumetric flask approximately two thirds full and mix. Carefully fill the flask to the mark etched on the neck of the flask. Use a wash bottle or medication dropper if necessary. Mix the solution wholly by using stoppering the flask securely and inverting it ten to twelve times.

A volumetric flask is used when it is imperative to be aware of each precisely and accurately the quantity of the solution that is being prepared. Like volumetric pipets, volumetric flasks come in distinctive sizes, depending on the extent of the answer being prepared.

Firmly stopper the flask and invert multiple times (> 10) to make certain the solution is nicely mixed and homogeneous. When working with a solute that releases warmth or gas all through dissolution, you ought to additionally pause and pull out the stopper once or twice. Use flasks for preparing options only.

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The grams of N2 are required to completely react with 3.03 grams of H2 for the following balanced chemical equation is 14 g.

We may calculate the number of moles of H2 that will be used by dividing the amount of H2 that will be utilised by its molar mass. We may multiply that number by the molar mass of N2 to get how many grammes we should use. We can divide that mole quantity by 3 to determine how many moles of N2 the reaction will consume.

In the reaction 1 mole of N2 react with  3 mole of H2 and give 2 mole of NH3

mass of H2 = 3.03g

No of moles of H2 = 3.03g/2 gmol-1

         = 1.51 mole

1.51 mole of H2 require N2 = (1/3)× 1.51 moles  

        = 0.50 mole N2

molar mass of N2  =28g/mol

Mass of N2 require   = 0.50mole ×28g/mol

    = 14g

Mass of N2 require = 14g.

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The answer is C. 14.0 grams of N2 are required to completely react with 3.03 grams of H2.

The balanced chemical equation is:

N2 + 3H2 -> 2NH3

From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

To find out how many grams of N2 are required to react with 3.03 grams of H2, we first need to convert 3.03 grams of H2 to moles:

moles of H2 = mass of H2 / molar mass of H2
moles of H2 = 3.03 / 2.016
moles of H2 = 1.505

Now, we can use the mole ratio from the balanced equation to find out how many moles of N2 are required to react with 1.505 moles of H2:

moles of N2 = (1.505 mol H2) / (3 mol H2/1 mol N2)
moles of N2 = 0.5017

Finally, we can convert moles of N2 to grams of N2:

mass of N2 = moles of N2 x molar mass of N2
mass of N2 = 0.5017 x 28.02
mass of N2 = 14.04

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Explain to the kids that since there is essentially no —which is required for fire—if the bag contains only pure carbon dioxide, the splint would burn out right away.

H2 - Hydrogen Pure hydrogen gas will burst into flames when a burning splint is added to it, making a popping sound. Oxygen (O2) A smouldering splint will rekindle when exposed to a sample of pure oxygen gas.

The flame goes out as a result of carbon dioxide replacing the oxygen it requires to burn (the effect). A popping sound is produced when a flame is near hydrogen because of how the gas burns.

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All of these solutions have the same concentration of Na⁺ (aq) at 3.0 moles for molar concentration.

The highest molar concentration of Na⁺ (aq) can be determined by calculating the moles of Na⁺ ions in each solution.

1. Identify the number of sodium ions (Na⁺) in each compound:
  - NaCl: 1 Na⁺ ion
  - NaC₂H₃O₂: 1 Na⁺ ion
  - Na₂SO₄: 2 Na⁺ ions
  - Na₃PO₄: 3 Na⁺ ions

2. Calculate the moles of Na⁺ ions in each aqueous solution:
  - 3.0 M NaCl: 3.0 M * 1 Na⁺ ion = 3.0 moles of Na⁺ ions
  - 3.0 M NaC₂H₃O₂: 3.0 M * 1 Na⁺ ion = 3.0 moles of Na⁺ ions
  - 1.5 M Na₂SO₄: 1.5 M * 2 Na⁺ ions = 3.0 moles of Na⁺ ions
  - 1.0 M Na₃PO₄: 1.0 M * 3 Na⁺ ions = 3.0 moles of Na⁺ ions

3. Compare the moles of Na⁺ ions in each solution to determine the highest concentration.

All of these solutions have the same concentration of Na⁺ (aq) at 3.0 moles.

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Though all the solutions have the same concentration of Na+ (aq), an aqueous solution of NaCl with 3.0 M has the highest molar concentration among the given solutions.

Explanation: To determine the molar concentration of Na+ (aq) in each solution, we need to consider the stoichiometry of the dissociation of each compound in water.

For sodium chloride (NaCl), it dissociates completely into Na+ and Cl- ions, so the molar concentration of Na+ (aq) is equal to the molar concentration of NaCl. Therefore, the molar concentration of Na+ (aq) in 3.0M NaCl is 3.0M.
For sodium acetate (NaC2H3O2), it dissociates into Na+ and C2H3O2- ions, but in a 1:1 ratio. So, the molar concentration of Na+ (aq) is half of the molar concentration of NaC2H3O2. Therefore, the molar concentration of Na+ (aq) in 3.0M NaC2H3O2 is 1.5M.
For sodium sulfate (Na2SO4), it dissociates into 2 Na+ ions and 1 SO4 2- ion. So, the molar concentration of Na+ (aq) is twice the molar concentration of Na2SO4. Therefore, the molar concentration of Na+ (aq) in 1.5M Na2SO4 is 3.0M.
For sodium phosphate (Na3PO4), it dissociates into 3 Na+ ions and 1 PO4 3- ion. So, the molar concentration of Na+ (aq) is three times the molar concentration of Na3PO4. Therefore, the molar concentration of Na+ (aq) in 1.0M Na3PO4 is 3.0M.

Therefore, the solution with the highest molar concentration of Na+ (aq) is 3.0M NaCl (sodium chloride).

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Answer:

using

V1/T1=V2/T2

make V2 subject of formula

V2= V1T2/T1

V2= 1.9L