Arrange the following from lowest to highest entropy: NO(g) , NO2(g), N2O4(g) , N2 (g) . Assume 1 mole of each at the same conditions.

Answer:

Molecular formula = C₄H₈

Explanation:

Given data:

Empirical formula = CH₂

Molar mass of compound = 56.12 g/mol

Molecular formula = ?

Solution:

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 12× 1 + 1.01 × 2= 14.02 g/mol

n = 56.12 / 14.02

n = 4

Molecular formula = n (empirical formula)

Molecular formula = 4 (CH₂)

Molecular formula = C₄H₈

Answer:

The sun's light is blocked by the moon.

Explanation:

During the eclipse, the moon rotates right in front of the sun, that's why the eclipse is so rare and only happens every four(?) years/

Answer:

H2SO4 + CO3 = H2CO3 + SO4

Explanation:

H2SO4 + CO3

This is the reaction of sulphuric acid and carbon trioxide.

Balanced reaction is;

H2SO4 + CO3 = H2CO3 + SO4

So they react to produce carbonic acid and sulfate.

The solution that will be a string one based on the information is D. NaC₂H₃O₂.

A medium that contains ions and is electrically conducting due to the movement of those ions but does not conduct electrons is called an electrolyte. This includes the majority of salts, acids, and bases that are soluble when dissolved in polar solvents like water.

A solution is a specific kind of homogeneous mixture made up of two or more substances that is used in chemistry. A homogeneous mixture of two or more substances in relative amounts that can be continuously varied up to what is referred to as the limit of solubility is known as such a mixture. A solute is a substance dissolved in a solvent. The word "solution" is most often used to refer to the liquid state of matter, but it is also possible for gases and solids to form solutions.

Strong electrolyte means conducts electricty in water i.e, soluble ionic compound. It ionises in water completely. The correct option is D.

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\(density = \frac{mass}{volume} \)

\(d = \frac{25.7}{22.6} = 1.13717 \: g/ml\)

\(pay \: attention \: to \: the \: unit \: used \: to \\ express \: the \: density \\ since \: the \: mass \: is \: given \: in \: grams \\ and \: volume \: in \: milliliter \: then \\ the \: density \: is \: expressed \: in \\ grams \: per \: \: milliliter\)

The rate constant for the reaction is either 7.14×10−3 s−1 or 2.50×10−3 s−1, depending on which rate was used to calculate it.

The rate of the reaction is given by the equation:

Rate = -k[A]

where k is the rate constant and [A] is the concentration of the reactant.

Rate at t=140 s:

Rate = (8.00×10−2 M - 0 M) / (140 s - 0 s)

= 5.71×10−4 M/s

Rate at t=400 s:

Rate = (4.00×10−2 M - 0 M) / (400 s - 0 s)

= 1.00×10−4 M/s

Since this is a zero-order reaction, the rate of the reaction is constant, and we can use either rate to calculate the rate constant:

k = Rate / [A]

Using the rate at t=140 s:

k = 5.71×10−4 M/s / 8.00×10−2 M = 7.14×10−3 s−1

Using the rate at t=400 s:

k = 1.00×10−4 M/s / 4.00×10−2 M

= 2.50×10−3 s−1

The rate constant for the reaction is either 7.14×10−3 s−1 or 2.50×10−3 s−1.

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The volume of hydrogen gas produced in the reaction of 73 grams of zinc and 73 grams of hydrochloric acid (under normal conditions) is approximately 22.4 liters.

To calculate the volume of hydrogen gas produced in the reaction of zinc and hydrochloric acid, we need to use the principles of stoichiometry and the ideal gas law.

First, let's write the balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl):

Zn + 2HCl →\(ZnCl_2\)+ H2

From the equation, we can see that one mole of zinc reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas. To determine the number of moles of zinc and hydrochloric acid, we need to convert the given masses into moles.

The molar mass of zinc (Zn) is approximately 65.38 g/mol, so 73 grams of zinc is equal to:

73 g Zn * (1 mol Zn / 65.38 g Zn) ≈ 1.116 mol Zn

Similarly, the molar mass of hydrochloric acid (HCl) is approximately 36.46 g/mol, so 73 grams of HCl is equal to:

73 g HCl * (1 mol HCl / 36.46 g HCl) ≈ 2.002 mol HCl

According to the balanced equation, the reaction produces one mole of hydrogen gas for every two moles of hydrochloric acid. Therefore, since we have 2.002 moles of HCl, we expect to produce half that amount, or approximately 1.001 moles of hydrogen gas.

To calculate the volume of hydrogen gas, we can use the ideal gas law, which states:

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 this case, we assume the reaction is conducted under normal conditions, which means a pressure of 1 atmosphere and a temperature of 273.15 Kelvin.

Rearranging the equation to solve for V, we have:

V = nRT / P

Substituting the values, we get:

V = (1.001 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm) ≈ 22.4 L

Therefore, the volume of hydrogen gas produced in the reaction is approximately 22.4 liters.

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Answer:

The d-block and transition metal are not interchangeable terms because the d-block elements are a subset of the transition metal elements. The transition metals are defined as the elements that have partially filled d orbitals, which includes the d-block elements as well as other elements that have partially filled d orbitals in other blocks, such as lanthanides and actinides. Therefore, while all d-block elements are transition metals, not all transition metals are d-block elements.

The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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Answer:

If n = 4, then the possible values of 1 and m depend on the equation or expression being used. Without more information, it is impossible to determine what the possible values of 1 and m might be. Can you please provide more context or information about the problem you are trying to solve?

Although sodium and phosphorus are both found in the same period of the periodic table, the nature of their oxides is different. Although P2O5 is acidic, Na2O is basic.

Metal oxides are basic, whereas non-metal oxides are acidic. The metallic character of the elements lessens as you move from left to right over time, while the non-metallic character grows.

Since the oxides in an element's higher oxidation state are more acidic than those in its lower oxidation state, dinitrogen pentoxide is the most acidic oxide compared to nitrogen dioxide and nitrogen tetraoxide.

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Answer:

39.1

Explanation:

i took the test.

Answer:

A precipitate will form when a freshly prepared aqueous carbonic acid solution is added to an aqueous solution of Calcium Hydroxide.

Explanation:

A precipitate will form when a freshly prepared aqueous carbonic acid

solution reacts with an aqueous solution of Calcium Hydroxide

Ca(OH)₂ + H₂CO₃ → CaCO₃ + 2 H₂O

This is an example of a neutralization reaction between the acid ( H₂CO₃)

and a base ( Ca(OH)₂ ) to form a precipitate known as Calcium

carbonate(CaCO₃).

The product which is Calcium carbonate(CaCO₃) is also referred to as calcite

or chalk.

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A centripetal force is experienced by any item traveling in a circle (or along a circular path).

A centripetal force is a force that causes a body to move in a circular motion. The centripetal force always acts orthogonally to the motion of the body and towards the stationary point of the path's instantaneous center of curvature.

Centripetal force is demonstrated by the rotation of the moon around the earth and the spinning of the top.

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Answer:

What Affects Reaction Rate?

The purpose of this lab was to see how temperature and particle size affects reaction rate. The first hypothesis is if you increase the temperature of a reaction, then the reaction rate will increase because particles experience more collisions at higher temperatures.The second hypothesis is if you decrease the particle size of a reactant, then the reaction rate will increase because more of the reactants’ molecules will contact each other. The independent variables are particle size and temperature. The dependent variable is reaction rate.

Materials

250 mL graduated cylinder

Thermometer

Water

Timer

Four 250 mL beakers

Seven 1,000 mg effervescent tablets

Two pieces of filter paper

600 mL beaker

Ice

Hot plate

Procedure

Step 1:Gather Materials

Variation of Temperature

Step 2:Measure the Reaction Rate at ≈ 20°C (Room Temperature)

a) Using a graduated cylinder, fill a 250 mL beaker with 200 mL of water.

b) Measure the temperature of the water and record it in the correct row of Table A.

c) Reset the timer. Start the timer as you place a full tablet into the beaker.

d) Record the reaction time on the Data Sheet in the correct row of Table A.

e) Compute the reaction rate to the nearest mg/L/sec. Record it in the last column of Table A. Measure the Reaction Rate at ≈ 40°C

Step 3:Repeat Step 2, heating the water to approximately 40°C using a hot plate during sub-step a. Measure the Reaction Rate at ≈ 65°C

Step 4:Repeat Step 2, heating the water to approximately 65°C using a hot plate during sub-step a. Measure the Reaction Rate at ≈ 5°C

Step 5:Repeat Step 2, chilling the water to approximately 5°C inside an ice bath during sub-step a. (To create an ice bath, place 100 mL of ice and 100 mL of water in a 600 mL beaker of ice water and wait until the temperature reaches approximately 5°C. To save time, you may wish to set up the ice bath, using an additional 250 mL beaker, while working on Step 4.)

Variation of Particle Size

Step 6:Measure the Reaction Rate for a Full Tablet

a) Using a graduated cylinder, fill a 250 mL beaker with 200 mL of water.

b) Reset the timer. Start the timer as you place the tablet in the beaker.

c) Record the reaction time on the Data Sheet in the appropriate row of Table B.

d) Compute the reaction rate to the nearest mg/L/sec. Record it in the last column of Table B.

Step 7:Measure the Reaction Rate for a Partially Broken Tablet

Repeat Step 6, but this time break the tablet into eight small pieces on a piece of filter paper. Make sure to place all of the pieces into the beaker at the same time.

Step 8:Measure the Reaction Rate for a Crushed Tablet

Repeat Step 6, but this time crush the tablet into tiny pieces on a piece of filter paper. Make sure to place all of the pieces into the beaker at the same time.

Step 9: Dispose of all samples according to your teacher’s directions.

Measured Reaction Temperature (°C)

Mass of Tablet (mg)

Volume of Water (L)

Reaction Time (s)

Reaction Rate (mg/L/s)

≈20°C

24

1,000

0.2

34.2

146.2

≈40°C

40

1,000

0.2

26.3

190.1

≈65°C

65

1,000

0.2

14.2

352.1

≈5°C

3

1,000

0.2

138.5

36.1

Relative Particle Size (Small, Medium, Large)

Mass of Tablet (mg)

Volume of Water (L)

Reaction Time (s)

Reaction Rate (mg/L/s)

Full Tablet

large

1,000

0.2

34.5

144.9

Broken Tablet

medium

1,000

0.2

28.9

173.0

Crushed Tablet

small

1,000

0.2

23.1

216.5

The data in the first table show that as the temperature increases the reaction time decreases and in turn the reaction rate increases. The data supported the hypothesis that as temperature increases reaction rate will also increase. The second table shows that as the particle size decreases the reaction time increases because there is more surface area when the particles are smaller. The data in the second table supported the second hypothesis that as particle size decreases the reaction rate will increase because there will be more contact in the molecules. Possible source of error would be an error in stopping the timer in time or chips in the tablets. To improve this lab it could be done with different types of reactions or different temperature or different particle sizes.

Explanation:

Answer:

The person above is amazing and gave us all the answers!!

Explanation:

Answer:

The Sun is a star.

Explanation:

Answer:

sound and sunlight are not considered matter but, the rest is matter

Explanation:

At equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

1: Write the balanced chemical equation:

\(C_O\)(g) + \(Cl_2\)(g) ⟶ \(C_OCl_2\)(g)

2: Set up an ICE table to track the changes in moles of the substances involved in the reaction.

Initial:

\(C_O\)(g) = 0.3500 mol

\(Cl_2\)(g) = 0.05500 mol

\(C_OCl_2\)(g) = 0 mol

Change:

\(C_O\)(g) = -x

\(Cl_2\)(g) = -x

\(C_OCl_2\)(g) = +x

Equilibrium:

\(C_O\)(g) = 0.3500 - x mol

\(Cl_2\)(g) = 0.05500 - x mol

\(C_OCl_2\)(g) = x mol

3: Write the expression for the equilibrium constant (Kc) using the concentrations of the species involved:

Kc = [\(C_OCl_2\)(g)] / [\(C_O\)(g)] * [\(Cl_2\)(g)]

4: Substitute the given equilibrium constant (Kc) value into the expression:

1.2 x \(10^3\) = x / (0.3500 - x) * (0.05500 - x)

5: Solve the equation for x. Rearrange the equation to obtain a quadratic equation:

1.2 x \(10^3\) * (0.3500 - x) * (0.05500 - x) = x

6: Simplify and solve the quadratic equation. This can be done by multiplying out the terms, rearranging the equation to standard quadratic form, and then using the quadratic formula.

7: After solving the quadratic equation, you will find two possible values for x. However, since the number of moles cannot be negative, we discard the negative solution.

8: The positive value of x represents the number of moles of \(Cl_2\)(g) at equilibrium. Substitute the value of x into the expression for \(Cl_2\)(g):

\(Cl_2\)(g) = 0.05500 - x

9: Calculate the value of \(Cl_2\)(g) at equilibrium:

\(Cl_2\)(g) = 0.05500 - x

\(Cl_2\)(g) = 0.05500 - (positive value of x)

10: Calculate the final value of \(Cl_2\) (g) at equilibrium to get the answer.

Therefore, at equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

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Answer:

1-water 2-Sun 3-Gravity

Answer:

A. Iron

Explanation

To graph the function g(x) = f(-x), you can start with the graph of f(x) and then reflect it about the y-axis.

To find the graph of the function g(x) = f(-x), we can start with the graph of the function f(x) and then reflect it about the y-axis.

If the graph of f(x) is symmetric with respect to the y-axis, meaning it is unchanged when reflected, then g(x) = f(-x) will have the same graph as f(x).

However, if the graph of f(x) is not symmetric with respect to the y-axis, then g(x) = f(-x) will be a reflection of f(x) about the y-axis.

In either case, the resulting graph of g(x) = f(-x) will be symmetric with respect to the y-axis.

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The Keq for the reaction will be 0.89.

The concentration of CO2 Will be 1.67M.

The equilibrium constant (Keq) for the reaction formulated as Keq = [H2O]^2 / ([H2]^2 [O2]), can be evaluated using the given concentrations of hydrogen and oxygen.

After applying both their respective quantities, we arrive at a calculated value of 0.89 for Keq. Moving on, to calculate the concentration of CO2 in the equation H2O (1) + CO2 (g) → H2CO3 (aq), where Keq = [H2CO3] / ([H2O] [CO₂]), having been supplied with a value of Keq equal to 0.15 and H2CO3 being at 0.25M, the rearranged expression reveals that [CO2] equals 1.67 M following basic algebraic substitution.

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The IUPAC name for each of the compounds would be:

A. 2,6-Dimethyl octane

B. Octane

IUPAC naming is a system of naming organic compounds according to the rules set up by the International Union of Pure and Applied Chemistry.

According to these rules:

Applying these rules to the structures in the image, the IUPAC names would be as follows:

A. 2,6-Dimethyl octane

B. Octane

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Answer:

sdfhioupsdfiuhikdfjsdfhsdksdflk";d089sdfojskdfk

pls give brainlyest

Explanation:

200 IQ intellectual right here ^^

There are approximately 6.022 × 10^23 atoms in 12 grams of Carbon-12 (12C).

The number of atoms in a given amount of a substance can be calculated using Avogadro's number, which represents the number of atoms or molecules in one mole of a substance. Avogadro's number is approximately 6.022 × 10^23.

Carbon-12 is a specific isotope of carbon, with an atomic mass of 12 atomic mass units (amu). One mole of Carbon-12 has a mass of 12 grams. Since one mole of any substance contains Avogadro's number of particles, in the case of Carbon-12, it contains 6.022 × 10^23 atoms.

Therefore, if we have 12 grams of Carbon-12, which is equal to one mole, we can conclude that there are approximately 6.022 × 10^23 atoms in this amount of Carbon-12.

In summary, 12 grams of Carbon-12 contains approximately 6.022 × 10^23 atoms. Avogadro's number allows us to relate the mass of a substance to the number of atoms or molecules it contains, providing a fundamental concept in chemistry and enabling us to quantify and understand the microscopic world of atoms and molecules.

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Answer:

116.6 lbs

Explanation:

There are 2.2 lbs per Kilogram of weight - and likewise 0.454 Kilograms per pound - but instead of dividing by .454 I multiplied the weight by 2.2 to get 116.6 pounds (of course you could round up and get 117 but 116.6 is a little more accurate).

The weight of the stone in pounds will be 116.8 pounds.

We have a large stone of which weighs 53 Kg.

We have to find its weight in pounds.

In 1 kilogram there are 2.2046 lbs.

According to the question -

Weight of stone in kilograms = 53 Kg

Assume that the weight of stone is equal to A kg. Then -

A = 53 Kg

Now, in order to convert A kg into pounds, we will multiply it by 2.2046.

Therefore, Weight of stone in kilograms is equivalent to = 53 x 2.2046 = 116.8 pounds.

Hence, the weight of the stone in pounds will be 116.8 pounds.

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Energy lost to condense = 803.4 kJ

Condensation of steam through 2 stages:

1. phase change(steam to water)

2. cool down(100 to 0 C)

1. phase change(condensation)

Lv==latent heat of vaporization for water=2260 J/g

\(\tt Q=300\times 2260=678000~J\)

2. cool down

c=specific heat for water=4.18 J/g C

\(\tt Q=300\times 4.18\times (100-0)=125400\)

Total heat =

\(\tt 678000+125400=803400~J\)

Answer:

answer is a

Explanation:

i took quiiiiiiz

Because they are autotrophs, plants make their own food. Water, sunshine, and carbon dioxide are converted into oxygen and simple sugars that the plant uses as fuel through the process of photosynthesis.

By using chemicals as their energy source rather than sunlight, bacteria produce food (glucose) through a process known as chemosynthesis. Where there is no sunshine, chemosynthesis takes place at hydrothermal vents and methane seeps in the deep ocean.

Chemosynthesis is the process by which bacteria or other living things create organic compounds utilising energy from chemical processes involving inorganic substances, usually without the use of sunlight. vegetation because of the lack of light.

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