What is the empirical formula of a compound that is composed of 60.94% carbon 15.36% hydrogen and 23.70% nitrogen

The properties of a natural resource that make it useful for humans as a material or energy source is the ability to convert mass into energy and vice versa.

The expression natural resources make reference to all types of matter and energy extracted from nature that can be used to produce goods and services.

Some examples of natural resources include for example irreversible resources such as fossil fuels (i.e., oil, or coal, gas, minerals such as metals, rocks, etc) as well as those based on the use of reversible energy such as eolic air energy, solar radiation or sunlight, soil and hydric resources or water.

Therefore, with this data, we can see that natural resources can be defined as any material and or energy obtained from nature that may be irreversible or reversibly used to produce goods and services.

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Answer:

932.27 i think

Explanation:

Explanation:

Given:

Molar mass of the compound = 133.60 g/mol

Volume of the solution = 471.4 mL

Molarity of the solution = 0.3 M.

The mass in grams of the compound = unknown

Note: Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters, i.e

1 mL = 0.001 L

So, 471.4 mL will be 0.001 x 471.4 = 0.4714 L

Therefore, the mole of the compound can be calculated as follows:

Answer:

Gases are more difficult to measure than liquids, because measured volumes are highly affected by temperature and pressure. Gas meters measure a defined volume, regardless of the pressurized quantity or quality of the gas flowing through the meter.

Explanation:

It is difficult to measure the volume of agas because measured volumes are highly affected by temperature and pressure.

The empirical formula of the compound is C₂H₁₆S₂N₃O.

The moles of each element is as follows::

For CO₂:

Carbon (C) has a molar mass of 12.01 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of C in CO₂ = 2.08 g / 12.01 g/mol = 0.173 moles

Moles of O in CO₂ = 2.08 g / 16.00 g/mol = 0.130 moles

For H₂O:

Hydrogen (H) has a molar mass of 1.01 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of H in H₂O = 1.28 g / 1.01 g/mol = 1.27 moles

Moles of O in H₂O = 1.28 g / 16.00 g/mol = 0.080 moles

For SO₃:

Sulfur (S) has a molar mass of 32.06 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of S in SO₃ = 4.77 g / 32.06 g/mol = 0.149 moles

Moles of O in SO₃ = 4.77 g / 16.00 g/mol = 0.298 moles

For HNO₃:

Hydrogen (H) has a molar mass of 1.01 g/mol.

Nitrogen (N) has a molar mass of 14.01 g/mol.

Oxygen (O) has a molar mass of 16.00 g/mol.

Moles of H in HNO₃ = 3.48 g / 1.01 g/mol = 3.45 moles

Moles of N in HNO₃ = 3.48 g / 14.01 g/mol = 0.248 moles

Moles of O in HNO₃ = 3.48 g / 16.00 g/mol = 0.217 moles

The simplest whole-number ratio of the elements will be:

Carbon: 0.173 moles / 0.080 moles ≈ 2.16

Hydrogen: 1.27 moles / 0.080 moles ≈ 15.88

Sulfur: 0.149 moles / 0.080 moles ≈ 1.86

Nitrogen: 0.248 moles / 0.080 moles ≈ 3.10

Oxygen: 0.080 moles / 0.080 moles = 1

Therefore, the empirical formula is C₂H₁₆S₂N₃O.

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The volume (in mL) of 0.100 M Na₂CO₃ needed to produce 1.00 g of CaCO₃ is 100 mL

First, we shall determine the mole in 1.00 g of CaCO₃. Details below:

Mole = mass / molar mass

Mole of CaCO₃ = 1 / 100.09

Mole of CaCO₃ = 0.01 mole

Next, we shall obtain the mole of Na₂CO₃. Details below:

Na₂CO₃ + CaCl₂ -> 2NaCl + CaCO₃

From the balanced equation above,

1 mole of CaCO₃ were obtained from 1 mole of Na₂CO₃

Therefore,

0.01 moles of CaCO₃ will also be obtain from 0.01 mole of Na₂CO₃

Finally, we shall determine the volume of Na₂CO₃ needed. Details below:

Volume = mole / molarity

Volume of Na₂CO₃ = 0.01 / 0.1

Volume of Na₂CO₃ = 0.1 L

Multiply by 1000 to express in mL

Volume of Na₂CO₃ = 0.1  1000 =

Volume of Na₂CO₃ = 100 mL

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Answer:

One mole of ideal gas occupies 22.4L volume at STP

Explanation:

So., we treat carbon dioxide an ideal gas:

3 mol x 22.4L •Mol-¹ ≈? ?L

The right answer is (KP)-1/2. Option A.

An expression of the equilibrium constant in terms of partial pressure is denoted as Kp. The equilibrium constant Kp is equal to the partial pressure of the product divided by the partial pressure of the reactant, and the partial pressure increases with a force equal to the coefficient of the substance in the equilibrium equation.

Kp is the equilibrium constant used to express equilibrium concentrations at atmospheric pressure and Kc is the equilibrium constant used to express equilibrium concentrations in molar terms. The relationship between Kp and Kc depends on the change in the number of moles of gaseous reactants and products. Kp has exactly the same form as Kc but partial pressure is used instead of concentration. Gases to the right of the formula are at the top of the print, and gases to the left are at the bottom.

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63.4 g of CH is needed to form 112 g of CH₂Cl₂.

Excess Cl₂ is assumed in the given reaction to ensure that all the CH available is consumed completely in the reaction, and there is no Cl₂ left over.

Let's assume x grams of CH is needed to form 112 g of CH₂Cl₂.

From the balanced equation of the second reaction, we can say that one mole of CH produces one mole of CH₂Cl₂.

Molar mass of CH₂Cl₂ = 12.01 + 2(1.01) + 2(35.45) = 84.93 g/mol

Number of moles of CH₂Cl₂ = 112 g / 84.93 g/mol = 1.318 mol

Since 65% yield is given for the second reaction, the actual amount of CH,CI produced will be 0.65 mol.

From the balanced equation of the first reaction, we can say that one mole of CH reacts with one mole of Cl₂ to produce one mole of CH,CI.

Since 82% yield is given for the first reaction, the actual amount of CH needed will be 0.65 / 0.82 = 0.793 mol.

Now, we can calculate the mass of CH needed as follows:

Mass of CH needed = 0.793 mol x 16.04 g/mol = 12.71 g

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The appropriate metric prefix depends on the exponent of the measurement.

The metric system is a universally acceptable means of measurement that makes use of the litre, gram, meter as units of volume, mass and length respectively.

The following are the most appropriate metric prefixes for the measurements;

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1. The IUPAC name of N is nitrogen.

2. Nitrogen dioxide

3.The IUPAC name of O is oxygen

4.The IUPAC name of OH is hydroxyl.

The IUPAC name of ·0 is a radical. It is commonly found in organic chemistry and plays an important role in many reactions.

IUPAC names for the given compounds are:1.1. N: Nitrogen

The IUPAC name of N is nitrogen.

It is a non-metal and belongs to group 15 in the periodic table. It has an electronic configuration of 1s2 2s2 2p3.1.2. NO2: Nitrogen dioxide

Explanation: NO2 is a chemical compound that is formed by the combination of nitrogen and oxygen. It is a reddish-brown gas that has a pungent odor.

The IUPAC name of NO2 is nitrogen dioxide.1.3. O: Oxygen

Explanation: The IUPAC name of O is oxygen.

It is a non-metal and belongs to group 16 in the periodic table. It has an electronic configuration of 1s2 2s2 2p4.

X: UnknownExplanation: No IUPAC name can be given to an unknown compound as the structure and composition are not known.

Y: Hydroxyl Explanation: The IUPAC name of OH is hydroxyl.

It is a functional group that is composed of an oxygen atom and a hydrogen atom (-OH). It is commonly found in alcohols and phenols. ·0: RadicalExplanation: A radical is a molecule or an ion that contains an unpaired electron.

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Note: The complete question is given below

Provide the IUPAC names for the following compounds:

\(CH_3CH_2CH(CH_3)CH_2CH_2CH_2CH_3\)

C6H5CH(CH3)2

H2NCH2CH2CH2CH2CH2NH2

CH3CH2CH2CH2CH2OH

CH3CH2CH2CHOHCH3

The final pressure of the gas, when its volume is increased from 3.4 L to 4.0 L at a constant temperature, is approximately 2.55 atm.

To determine the final pressure of a gas when its volume is changed while its amount and temperature remain constant, we can use Boyle's law. Boyle's law states that the pressure of a gas is inversely proportional to its volume at constant temperature.

Mathematically, Boyle's law can be expressed as:

P1 * V1 = P2 * V2

Where:

P1 = Initial pressure

V1 = Initial volume

P2 = Final pressure

V2 = Final volume

In this case, we are given:

P1 = 3.0 atm

V1 = 3.4 L

V2 = 4.0 L

We need to find P2, the final pressure.

Rearranging the formula, we have:

P2 = (P1 * V1) / V2

Substituting the values we have:

P2 = (3.0 atm * 3.4 L) / 4.0 L

P2 = 10.2 atm / 4.0 L

Calculating P2:

P2 = 2.55 atm

Therefore, the final pressure of the gas, when its volume is increased from 3.4 L to 4.0 L at a constant temperature, is approximately 2.55 atm.

Boyle's law demonstrates the inverse relationship between pressure and volume for a gas at a constant temperature. As the volume increases, the pressure decreases, and vice versa, as long as the amount and temperature of the gas remain constant.

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A neutral particle of an atom is located at the nucleus.

Electrons are those negatively charged particles that are found revolving round an atom.

Electron cloud is defined as the model of an atom in which the atom consists of a small but massive nucleus surrounded by a cloud of rapidly moving electrons.

A neutral particle is a nucleus because the nucleus of an atom is made up of neutral particles found within the nucleus of an atom which doesn't contain any charges.

An atom is defined as the smallest and indivisible part of an element which can take part in a chemical reaction.

An atom is made up of nucleus which consists of the proton (which is positively charged) and the neutron (which bears no charges at all).

Therefore, a neutral particle can be said to be an electron cloud.

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Explanation:

Hope this somewhat helped:

a. The equation for the reaction of weak acid, C6H5CO2H, with water is: C6H5CO2H + H2O <=> C6H5CO2^- + H3O^+. We can assume that all the C6H5CO2H has reacted with water to form C6H5CO2^- and H3O^+.

The pH is defined as -log(H30^+ concentration), where H30^+ is the hydronium ion, a measure of the acidic strength of the solution. At the equivalence point, all the C6H5CO2H has reacted with the NaOH to form C6H5CO2^- and Na^+. The stoichiometry of the reaction can be determined by balanced the reaction using the information given.

C6H5CO2H + NaOH <=> C6H5CO2^- + Na^+

This reaction tells us that one mole of C6H5CO2H will react with one mole of NaOH to form one mole of C6H5CO2^- and one mole of Na^+.

There are many ways  in which it can be ensured that chemicals do not get spilled .

How quickly a chemical spill or leak can cause a serious accident or disaster is shocking. Chemical production, storage, and transportation all present a multitude of potential accident sites due to their volatile nature. Chemical safety in the workplace can be greatly increased by putting in place a strong emergency response plan and well-established spill prevention procedures.

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Answer:

Explanation:

Which of the following is most likely to be ductile?

a. Metal
b. Nonmetal
c. Metalloid
d. Gas

Answer: a. MetalMetal

The coefficients needed to balance the equation would be 2, 7, 4, and 6 respectively.

This requires that the number of atoms of each element is the same both on the reactant and product's sides.

Hence, the balanced equation of the reaction becomes:

\(2C_2 H_6 O+ 7O_2 --- > 4CO_2 +6H_2 O\)

Thus, the coefficient for each species would be 2, 7, 4, and 6 respectively.

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The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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Iodine, Mn, Helium and Copper are classified as Single atoms. Fluorine, No2 and Br2 can be classified as Diatomic molecules. CoCI2, Ba(OH)2, CdCI2 and K2CrO4 are classified as Formula units. C3H8 come under Molecules category.

A diatomic molecule is made up of two atoms of the same or different elements chemically bonded together. So, we can say that diatomic molecule has two atoms.

A formula unit refers to the smallest representative unit of an ionic compound that can exist independently and still retain the chemical and physical properties of the compound. It is the simplest whole number ratio of ions that make up an ionic compound, and it is the smallest unit of an ionic compound that retains its chemical properties.

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Answer:

18.57 is the relative mass

Explanation:

To find relative Atomic mass we need to multiply respective isotope percentage with its mass then add all mass× percentage and divided by 100

Solution

73×18+ 12×19 + 15×21 ÷ 100

1314+228+315÷100 = 1857÷100 = 18.57

Answer:

C

Explanation:

it should be C

Considering the Hess's Law, the enthalpy change for the reaction is -84.4 kJ.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

2 C (graphite) + 3 H₂(g) → C₂H₆(g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: C₂H₆(g) + \(\frac{7}{2}\) O₂(g) → 2 CO₂(g) + 3 H₂O(l) ; ΔH° = –1560 kJ

Equation 2:  H₂(g) + \(\frac{1}{2}\) O₂(g) → H₂O(l) ; ΔH° = –285.8 kJ

Equation 3: C(graphite) + O₂(g) → CO₂(g) ; ΔH° = –393.5 kJ

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 2 moles of C(graphite) on reactant side and it is present in third equation. In this case it is necessary to multiply it by 2 to obtain the necessary amount. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

Now, you need 3 moles of H₂(g) on reactant side and it is present in second equation. In this case it is necessary to multiply it by 3 to obtain the necessary amount and the variation of enthalpy also is multiplied by 3.

Finally, 1 mole of C₂H₆(g) must be a product and is present in the first equation. Since this equation has 1 mole of C₂H₆(g) on the reactant side, it is necessary to locate the C₂H₆(g) on the reactant side (invert it). When an equation is inverted, the sign of delta H also changes.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1:  2 CO₂(g) + 3 H₂O(l) → C₂H₆(g) + \(\frac{7}{2}\) O₂(g); ΔH° = 1560 kJ

Equation 2:  3 H₂(g) + \(\frac{3}{2}\) O₂(g) → 3 H₂O(l) ; ΔH° = –857.4 kJ

Equation 3: 2 C(graphite) + 2 O₂(g) → 2 CO₂(g) ; ΔH° = –787 kJ

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

2 C (graphite) + 3 H₂(g) → C₂H₆(g)    ΔH= -84.4 kJ

Finally, the enthalpy change for the reaction is -84.4 kJ.

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By using the ideal gas law and Avogadro's number, we find that there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

To determine the number of molecules of CO2 in 2.5 L at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's number.

Avogadro's number (N_A) is a fundamental constant representing the number of particles (atoms, molecules, ions) in one mole of substance. Its value is approximately 6.022 × 10^23 particles/mol.

STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (1 atm).

First, we need to convert the volume from liters to moles of CO2. To do this, we use the ideal gas law equation:

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.

Since we have STP conditions, we can substitute the values:

(1 atm) × (2.5 L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K).

Simplifying the equation:

2.5 = n × 22.4149.

Solving for n (the number of moles):

n = 2.5 / 22.4149 ≈ 0.1116 moles.

Next, we can calculate the number of molecules using Avogadro's number:

Number of molecules = n × N_A.

Number of molecules = 0.1116 moles × (6.022 × 10^23 particles/mol).

Number of molecules ≈ 6.72 × 10^22 molecules.

Therefore, there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

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Answer:

d=m/

Explanation:

d is density, m is mass, v is volume

Given: m =10.4g, v=12.2mL

substituting in equation,

d=10.4/ 12.2

d=0.8524g/mL

To learn more about density:

The density of the liquid is 0.852 g/mL.

To calculate the density of the liquid, we need to use the formula:

Density = Mass / Volume

Given that the mass of the liquid is 10.4 g and the volume is 12.2 mL, we can substitute these values into the formula:

Density = 10.4 g / 12.2 mL

Simplifying this expression, we find:

Density = 0.852 g/mL

Density is a physical property of a substance and is defined as the amount of mass per unit volume. In this case, the density tells us that for every milliliter of the liquid, there is 0.852 grams of mass. The units of grams per milliliter (g/mL) indicate that the density is a ratio of mass to volume.It is important to note that the density of a substance can vary with temperature, so this value is only valid under the conditions at which the measurement was made. Additionally, the density can provide valuable information about the identity of a substance, as different substances have different densities.

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When iron will react with oxygen it will form iron oxide. When 4.48 grams of iron is mixed with an excess oxygen 21.6 grams of iron oxide will produce.

When 4.8 grams of iron is combined with oxygen it will form iron oxide.

4.8 g O₂

= 4.8/32

= 0.15 mol.

0.15 mol O₂ produces 2 x 0.15 = 0.3 mol FeO.

Multiplying this by the molar mass of FeO  (72) produces 21.6 grams.

Theoretical yield = 21.6 grams % yield

= 18.4/21.6 = 85%

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Exothermic refers to chemical interactions that aerobic respiration. Combustion reactions release higher energy. Endothermic refers to atoms and molecules that either use or absorb reactive power.

A chemical process known as an endothermic releases energy as heat or light. It is an endothermic reaction's opposite. Chemical equation expressed as reactants + products + energy. An reaction mechanism is one in which electricity is given off as light or warmth.

A response is deemed to be exothermic if it produces heat while also undergoing a net decrease in basic enthalpy change. Samples include those type of combustion, iron rust, including water froze. Exothermic processes are those that discharge heat and energy into the surroundings.

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Batteries, natural gas, and coal are all chemical energy