the mass of carbon, nitrogen, and hydrogen are 12 amu, 16 amu, and 1 amu. what is the molecular weight of this molecule?

Geologists use rock samples including sedimentary rocks as direct evidence and seismic waves as indirect evidence.

They are the types of rocks which are formed as a result of accumulation of minerals or organic particles at the surface of the earth.Sedimentation is the name given to a collective process which cause the particles to settle.

The study of sedimentary rocks is useful in civil engineering and as important sources of coal, fossil fuels and ores.The color of sedimentary rock is often determined by iron content and sometimes due to the presence of organic matter.

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Gold cannot replace Magnesium from a compound as gold is placed lowest in the reactivity series of metals. It won't even react with the compound let alone replace it. Gold has great ductile strength and malleability.

 Features of Reactivity Series of Metals

Therefore Gold cannot replace Magnesium from a compound as it is the least reactive metal and therefore, comes lower than Magnesium in the reactivity series

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Answer:

the red arrow shows that the change in enthalpy is positive

From the energy diagram, it is clear that, the enthalpy of the reaction is positive and it is an endothermic reaction.

An endothermic reaction is a chemical reaction that absorbs heat from its surroundings, resulting in a decrease in temperature of the surroundings. In other words, energy is taken in by the reaction from the environment.

During an endothermic reaction, the reactants absorb energy in the form of heat, light, or electricity to break the bonds between their atoms and form new bonds to create products.

The amount of energy absorbed by the reactants is greater than the energy released by the products. This energy difference is usually expressed as the enthalpy change (∆H) of the reaction, which is positive for endothermic reactions.

Here, the energy of products is higher than that of reactants and hence option A is correct.

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Explanation: its moles

Answer:

it is becuse the students used wrong apparatus

The biggest error in this investigation is the use of inappropriate types of equipment.

The types of equipment used in the given setup are inappropriate. Suggested improvements in the apparatus to increase the accuracy are given below:

1. The use of a polystyrene cup in the laboratory for any reaction is inappropriate. As a slight increase in temperature causes the decomposition of the chemicals in the cup as well.

2. The use of a thermometer stand is a must. Without using the stand, the tip of the thermometer touches the lid of the container, which causes inaccuracies in the result.

Hence, all the suggestions are listed above.

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Alkyl bromides are created when an alkene and HBr interact in the presence of factors such as heat, sunshine, or peroxides. A radical intermediate is used in this reaction to progress.

What is Alkene?

Alkenes, and ethene in particular, play a crucial role in the chemical industry. They are created during the cracking of the alkanes but are not present in significant amounts in crude oil. Like all hydrocarbons, alkenes burn in the presence of air to produce carbon dioxide and water. Ethene reacts violently in oxygen, making it ineffective as a fuel. Additionally, the alkenes are too useful for the production of plastics and many other compounds to be used as fuels in the chemical industry.

Compound A reacts with HBr to produce 2-bromohexane as a byproduct. 1-Bromohexane is created when component an interacts with HBr and peroxide.

In the absence of light, the carbocation intermediate is used to add HBr. In the presence of light, the radical intermediate is used to add HBr.

The result of component B reacting with HBr is 1-bromo-1-methylcyclohexane. 1-Bromo-2-methylcyclohexane is produced via the reaction of component b with HBr and peroxide.

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Implementing a new system into an organization module by module is called a Phased Approach.

The "phased approach," as the name implies, is a project planning strategy in which anything new, such as a software solution, is introduced in stages rather than all at once.

Rather than installing and rolling out new software across an entire organization at the same time, older systems/methodologies are gradually replaced. Phased implementation necessitates extensive project planning to determine which software functions should go live first and for which departments.

This process allows you to keep the most critical departments operational while others transition to the new system. However, phasing may not be necessary for every company or project. As software and technology become more complex, however, the phased approach reduces risk and often makes the most business sense.

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The engineers fitted a multiple linear regression model to analyze the effect of variables on the efficiency of the ammonia oxidation process, assessed significance levels, and recommended a final model.

Engineers conducted an investigation on the efficiency of a process that oxidizes ammonia to nitric acid by studying the effect of three variables: Airflow, Water temperature, and Acid concentration.

They collected data on stack loss, which represents the percentage of unabsorbed ammonia escaping from the absorption tower. Using R, they fitted a multiple linear regression model with polynomial terms and interactions for the predictors.

They analyzed the model summary table to determine the mean square error (MSE), R², and adjusted R².

They also assessed the significance of variables/terms at a 0.1 significance level, tested the joint significance of interaction terms, and evaluated the joint significance of quadratic terms.

Based on the results, they recommended a final model containing only significant variables/terms at a significance level of 0.05.

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Answer:

1) A

2) P

3) B

4) M

5) I

6) C

7) H

8) J

9) F

10) L

11) N

12) G

13) O

14) K

15) D

Here ya go friend, these should be correct, for the most part.

In this reaction, oxygen can take on two electrons while calcium can lose two electrons. As a result, calcium provides oxygen with two valence electrons and creates an ionic connection with it.

The four heme groups in haemoglobin include four iron atoms that act as binding sites for oxygen molecules. A Hb tetramer can therefore bind four oxygen molecules. According to the molecular weight of Hb, 1.39 ml of oxygen may mix with 1 g of Hb.

To create Ca2+ and O2- ions, respectively, each calcium atom really loses two electrons and each O atom in O2.

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This statement is true.

The dry adiabatic lapse rate is the average temperature dereses with heigh of 6.5 degree celsius.

What is adiabatic lapse?

The rate of decrease of temperature experienced by a parcel of air when it is lifted in the atmosphere under the restriction that it cannot exchange heat with its environment. For parcels that remain unsaturated during lifting, the (dry adiabatic) lapse rate is 9.8°C per kilometer.

it averages about 6.5 °C per kilometre (18.8 °F per mile) in the lower atmosphere (troposphere). It differs from the adiabatic lapse rate, which involves temperature changes due to the rising.

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Answer:

Percent composition is calculated from a molecular formula by dividing the mass of a single element in one mole of a compound by the mass of one mole of the entire compound. This value is presented as a percentage.

Explanation:

1. To calculate the number of nanograms equivalent to 0.0078 mg, you need to multiply 0.0078 mg by the conversion factor of 1,000,000 ng/mg. The result is 7,800 nanograms (ng). 2. To convert 300 decigrams (dg) to micrograms (μg), you need to multiply 300 dg by the conversion factor of 100 μg/dg. The result is 3,000 micrograms (μg).

1. To calculate the number of nanograms equivalent to 0.0078 mg, conversion factors and the relationship between milligrams and nanograms need to be used. Direct calculation from milligrams to nanograms is not possible without considering the appropriate conversion factors.

To convert milligrams to nanograms, we need to consider the conversion factor: 1 milligram (mg) is equal to 1,000,000 nanograms (ng). By multiplying 0.0078 mg by the conversion factor (1,000,000 ng/mg), we can determine the equivalent value in nanograms.

0.0078 mg is equal to 7,800 nanograms (ng). The conversion from milligrams to nanograms requires the use of appropriate conversion factors, as the units differ by six orders of magnitude. It is essential to employ the correct conversion factors when converting between different units of measurement.

2. 300 decigrams (dg) is equal to 3,000 micrograms (μg).

To convert decigrams to micrograms, we need to consider the conversion factor: 1 decigram (dg) is equal to 100 micrograms (μg). By multiplying 300 dg by the conversion factor (100 μg/dg), we can determine the equivalent value in micrograms.

300 decigrams is equal to 3,000 micrograms. The conversion from decigrams to micrograms requires the use of the appropriate conversion factor, where decigrams are multiplied by 100 to obtain micrograms. Conversion factors play a crucial role in accurately converting between different units of measurement.

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Answer:

4.08

Explanation:

The atomic mass of an element is not solely determined by the most abundant isotope but rather takes into account the relative abundance of all naturally occurring isotopes. Although the relative abundance of the 180 isotope of hafnium (Hf) may be the highest, it does not necessarily mean that the atomic mass will be close to 180.

Isotopes are variants of an element that have different numbers of neutrons in their atomic nuclei. The atomic mass of an element is calculated as the weighted average of the masses of its isotopes, taking into account their relative abundances. So, even if the 180 isotope is the most abundant, if other isotopes with lower masses are also present, they will contribute to lowering the overall atomic mass.

In the case of hafnium, it has multiple isotopes, including 174Hf, 176Hf, 177Hf, 178Hf, 179Hf, and 180Hf, with varying abundances. The atomic mass of hafnium is calculated by considering the masses and relative abundances of all these isotopes, resulting in an atomic mass that may differ from the mass of any single isotope. Therefore, the atomic mass of hafnium is not close to 180, despite the high relative abundance of the 180 isotope.

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2Al + 3Cl₂ → 2AlCl₃

mol Al = 2/3 x 1.25 = 0.83

mass Al = 0.83 x 27 g/mol = 22.41 g

According to the stoichiometry 21.66 grams of aluminum will react fully with 1.25 moles Cl₂.

Stoichiometry is the determination of proportions of elements or compounds in a chemical reaction. The related relations are based on law of conservation of mass and law of combining weights and volumes.

Stoichiometry is used in quantitative analysis for measuring concentrations of substances present in the sample.It is useful in balancing chemical reactions.

According to the chemical reaction between aluminium and chlorine, 2 moles of aluminium reacts with 3 moles of chlorine , thus,1.25 moles of chlorine will react 1.25×2/3=0.83 moles which is mass of aluminium =0.83×26=21.66 g.

Thus, 21.66 grams of aluminum will react fully with 1.25 moles Cl₂.

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The enthalpy change needed for one mole of a substance to freeze is the opposite of the enthalpy change for melting, but with the same magnitude. Therefore, the enthalpy change for freezing would be ΔH = +79 kJ/mol. Option B.

When a substance melts, it absorbs heat from its surroundings, which leads to an increase in the enthalpy of the system. This increase in enthalpy is represented by a negative value for the enthalpy change, indicating an endothermic process. In this case, the given enthalpy change for melting is ΔH = -79 kJ/mol.

For the substance to freeze, the reverse process occurs. Heat is released from the substance, causing it to transition from the liquid phase to the solid phase. This release of heat results in a decrease in the enthalpy of the system. The magnitude of the enthalpy change for freezing would be the same as the enthalpy change for melting but with an opposite sign. Thus, the enthalpy change for freezing would be ΔH = +79 kJ/mol.

Therefore, the correct answer is (b) ΔH = +79 kJ/mol. It represents the enthalpy change needed for one mole of the substance to freeze. Option B is correct.

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Given the equation :

CH3OH +Cl2 →← 2 CH3CL + 2OH ^-1

at equilibrium concentration, we have:

(1.5 -2x )/ 5L + (1.0 -x )/5L =2x /5L + 2x /5L

→ (1.5-2x+ 1 -x )/ 5L = 4x /5L

→(0.5 -3x)/5L = 4x/5L (5L in the Left hand side cancels 5L inthe right hand side)

→ 0.5 = 4x +3x

0.5 = 7x

x = 0.5/7

x = 0.071

( Extra note : Now lets test and see if this balances ):

LHS : (1.5 -2x )/ 5L + (1.0 -x )/5L = 0.5 -3x)/5L

= 0.5 - 3(0.071)/5

= 0.287 /5 = 0.05

RHS :4(x) /5 = 4(0.071) /5

= 0.28/5 = 0.05

at equlibrium , LHS = RHS therefore our x value is 0.07

The mass of the water that is formed by the oxygen is 216 g.

We know that we would have to look at the reaction equation and this would give us an idea on how to be able to use the stoichiometry of the reaction equation so as to solve the problem that we have at hand here.

Then we have;

Equation of the reaction; \(2H_{2} +O_{2} --- > 2H_{2}O\)

Number of moles of the oxygen = 192 grams /32 g/mol

= 6 moles

If 1 mole of oxygen forms 2 moles of water

6 moles of oxygen would form = 6 * 2/1

= 12 moles of water

Mass of the water = 12 moles * 18 g/mol

= 216 g

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The mass of ammonia gas used in kg would be  1.02 kg..

Nitric (V) acid and ammonia gas react to form fertilizer according to the following balanced chemical equation:

\(HNO_3 + NH_3 -- > NH_4NO_3\)

From the equation, the ratio of the ammonium nitrate produced to the ammonia gas used is 1:1.

The molar mass of \(NH_4NO_3\) = 14 + 4 + 14 + 48 = 80 g/mol

Mole of 4800 kg  \(NH_4NO_3\) = 4800/80 = 60 mol

Thus, the equivalent amount of ammonia that needs to react will also be 60 mol.

Molar mass of amonnia = 14 + 3 = 17 g/mol

Mass of 60 mol ammonia = 60 x 17

                                           = 1020 grams

1020 grams = 1020/1000

                    = 1.02 kg

In other words, the mass of ammonia gas used would be 1.02 kg.

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the number of samples required so that the reported mean has a 90% confidence level is 72.

A lagoon waste pit has the following historical data for the barium concentration based on a simple random sampling (n = 4): 86, 90, 98, 104 mg/kg (the lower two thirds of lagoon). The regulatory threshold for barium is 100 mg/kg. The waste on this site was categorized to be hazardous, and therefore a more thorough sampling plan is needed. We have to determine the number of samples required so that the reported mean has a 90% confidence level.

To determine the number of samples required so that the reported mean has a 90% confidence level, we need to use the formula for the sample size.

n = [(z^2 * σ^2)/E^2]

Where n is the sample size, z is the standard normal value for the desired level of confidence, σ is the standard deviation, and E is the maximum error of estimate that is allowed.

Given,The lower two-thirds of the lagoon contains 86, 90, 98, 104 mg/kg barium concentration

The standard deviation is given by the formula as follows:

σ = √[(∑(x - μ)^2)/n]

where σ is the standard deviation, x is the individual data points, μ is the mean, and n is the sample size.

μ = [(86 + 90 + 98 + 104)/4]

= 94σ = √[((86 - 94)^2 + (90 - 94)^2 + (98 - 94)^2 + (104 - 94)^2)/4]

= 6.46E

is the maximum error of estimate that is allowed which is 2 mg/kg (100-98)z for a 90% confidence level is 1.645Putting all the values in the formula, we get;

n = [(z^2 * σ^2)/E^2]= [(1.645^2 * 6.46^2)/2^2]

= 71.8Since we cannot have 0.8 of a sample, we need to round the answer up, the required number of samples is 72.

Thus, the number of samples required so that the reported mean has a 90% confidence level is 72.

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The correct matches are:

(i) Acid chlorides and esters experience primarily dipole-dipole interactions between their molecules.

(ii) 1º and 2º amides experience primarily hydrogen bonding between their molecules.

(iii) Acid chlorides, esters, and 3º amides have lower boiling points than other carboxylic acid derivatives of similar size and shape.

(iv) 1º and 2º amides have the highest boiling points among other carboxylic acid derivatives of similar size and shape.

i.

Acid chlorides and esters experience primarily dipole-dipole interactions between their molecules due to the presence of polar bonds within the molecules.

ii.

1º and 2º amides experience primarily hydrogen bonding between their molecules, as they contain the necessary hydrogen and oxygen or nitrogen atoms for hydrogen bonding.

iii.

Acid chlorides, esters, and 3º amides have lower boiling points than other carboxylic acid derivatives of similar size and shape. This is due to the absence of hydrogen bonding in these compounds, resulting in weaker intermolecular forces.

iv.

1º and 2º amides have the highest boiling points among other carboxylic acid derivatives of similar size and shape. This is because they can form extensive hydrogen bonding between their molecules, leading to stronger intermolecular forces and higher boiling points.

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Answer: 80000 micrometers

hope i helped

Ionic Strength (µ) is a unit to define the overall effect of all ions that are present in a solution. It is given by the sum of the products of the concentrations of the ions (raised to the power of their stoichiometric coefficients) and the square of their charge numbers.

Let's calculate the ionic strength for each of the given solutions. A solution of 0.00525 M FeSO3.Assuming complete dissociation of FeSO3, FeSO3 → Fe³⁺ + 3SO₃²⁻(molar concentration of Fe³⁺) = 0.00525 M(molar concentration of SO₃²⁻) = 3 × 0.00525 M= 0.01575 M The total number of moles in 1 L of solution = 0.00525 + 0.01575 = 0.021 mole. Ionic strength of the solution µ = ½ [∑(Cazᵢ)²]∑(Cᵢzᵢ)² = (0.00525 × 3)² + (0.01575 × 2)² = 0.12069µ = ½ × 0.12069 = 0.06035A solution of 0.00106 M LaBr3.Assuming complete dissociation of LaBr3, LaBr3 → La³⁺ + 3Br⁻(molar concentration of La³⁺) = 0.00106 M(molar concentration of Br⁻) = 3 × 0.00106 M= 0.00318 MThe total number of moles in 1 L of solution = 0.00106 + 0.00318 = 0.00424 mole.Ionic strength of the solution µ = ½ [∑(Cᵢzᵢ)²]∑(Cᵢzᵢ)² = (0.00106 × 3)² + (0.00318 × (-1))² = 0.01107µ = ½ × 0.01107 = 0.00554A solution of 0.000879 M CaBr2 and 0.000666 M NaBr.Assuming complete dissociation of CaBr2, CaBr2 → Ca²⁺ + 2Br⁻(molar concentration of Ca²⁺) = 0.000879 M(molar concentration of Br⁻) = 2 × 0.000879 M= 0.001758 MThe total number of moles of Br⁻ ions in 1 L of solution = 0.001758 + 0.000666 = 0.002424 mole.Ionic strength of the solution µ = ½ [∑(Cᵢzᵢ)²]∑(Cᵢzᵢ)² = (0.000879 × 2)² + (0.000666 × (-1))² = 0.00282µ = ½ × 0.00282 = 0.00141The ionic strength for each of the given solutions are as follows:For a solution of 0.00525 M FeSO3, µ = 0.06035.For a solution of 0.00106 M LaBr3, µ = 0.00554.For a solution of 0.000879 M CaBr2 and 0.000666 M NaBr, µ = 0.00141.

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Answer:

The third choice

Explanation:

The law of conservation of mass tells us that the mass of the reactants and products in a chemical reaction will be equal to one another. Equations follow the law of conservation of mass when they are balanced, which means that they have the same number and type of atoms on each side.

When individual atoms are counted, the third choice has the same number and type of atoms on each side: 1 gallium, three cesium and three fluorine.

Answer:

If the reaction requires energy, it is often indicated above the arrow. A capital Greek letter delta (Δ) is written on top of the reaction arrow to show that energy in the form of heat is added to the reaction; hv is written if the energy is added in the form of light.

Answer:

it means that to get the product, the reactants were heated

The chemical reaction that we observe with the change in absorbance measured by the spectrophotometer is the oxidation of guaiacol by hydrogen peroxide (H2O2).

This chemical reaction is related to the activity of turnip peroxidase in that it is catalyzed by this enzyme. Turnip peroxidase is an enzyme that catalyzes the oxidation of a wide range of substrates, including guaiacol. Guaiacol, in particular, is a colorless compound that, upon oxidation, turns brown in the presence of peroxidase.Turnip peroxidase catalyzes the oxidation of H2O2 in the presence of guaiacol, resulting in the formation of a brown-colored product. This reaction results in a change in absorbance that can be measured using a spectrophotometer. Thus, the activity of turnip peroxidase is related to the rate of the oxidation of guaiacol by H2O2.

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