Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that 92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.

Answer:9.84x 10^-31m

Explanation: 9.84x 10^-31m

Heisenberg uncertainty principle states that  says that the position and momentum of a particle cannot be known at the same time and accurately. and that the values  of position and momentum all times must be greater than h/4π.

Using the formula for Heisenberg Uncertainty principle

Δx. Δp ≥ h / 4π

=Δx. m ΔV ≥ h / 4π

where h = Planck’s constant = 6.62607004 × 10-34 m2 kg / s

Δx is the uncertainty in position

Δp is the uncertainty in momentum

m = mass

Δv=is the uncertainty in velocity

Given,

v = 0.80m/s,

m = 0.67 g = 0.67 g / 1000

= 0.00067 kg

h = 6.62607004 × 10-34 m2 kg / s

uncertainty in the velocity is 0.1 m/s.

Δv = 0.80×0.1 = 0.080m/s

Δx. m ΔV ≥ h / 4π

Δx  ≥ h / 4πm ΔV

6.62607004 × 10-34 / 4 π x 0.08 x 0.00067

=6.62607004 × 10-34 / 4 x3.14  5.3 x 10^-5 = 9.84x 10^-31m

The general form of a neutralization reaction is HF + FrOH → FrF + H₂O

We refer to this as a neutralisation reaction. Only this reaction, which produces NaCl and water as products, is a neutralisation reaction since it involves HCl and NaOH. The resulting response is listed below: NaCl(aq) + H₂O = HCl(aq) + NaOH(aq) (l)

The interaction of H⁺ ions and OH⁻ ions produces water in a neutralisation reaction, which occurs when an acid and a base combine to make water and a salt. The neutralisation of a strong acid and strong base yields a pH of 7.

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The energy of the wave can be obtained as 6.6 * 10^-16 J.

We know that the energy of a wave has to do with the energy that is possessed by the wave and we can be able to obtain this energy by the means of the equation of Max Plank.

We know that we have the following equation;

E = hf

h = Plank's constant

f = Frequency of the wave

Then we have;

E = 6.6 * 10^-34 Js * 1 * 10^18 Hz

E = 6.6 * 10^-16 J

The radio waves have an energy of about 6.6 * 10^-16 J.

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Answer:

A liquid of unknown composition: Unknown liquid

A liquid of known composition: Known liquid

A solid-liquid mixture of unknown composition: Unknown solid-liquid mixture

A solid of known composition: Known solid

Molecular mass of  9.3 x 10^ 24 molecules of glucose  = 1675.6 x 10^ 24

It is given that the mass of glucose is 180.18 g/mol
so if we have to calculate the mass of 9.3x 10^ 24 molecules of glucose
we will apply a simple unitary method i.e,

9.3 x 10^ 24 *  180.18 g/mol = 1675.67 x 10^ 24
To calculate the molecular mass of a molecule, multiply the subscript (number of atoms) by the atomic mass of each element in the molecule and add those masses together.
Remember that - To determine the compound's molecular mass in grams per mole, use the molecular formula.
Divide the supplied mass by the molar mass of the chemical to convert it to moles.
By dividing the number of moles by Avogadro's number, you may convert from moles to molecules.

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(a) The hazard quotient is:

HQ = Intake / RfD = 0.0002 mg/day / 0.02 mg/kg/day = 0.01

(b) The cancer risk is 1 in 10,000.

(c) The cancer risk is 1 in 1,000.

Hazard quotient (HQ) is a measure used in risk assessment to evaluate the potential health risk posed by exposure to a chemical or other hazard. It is calculated as the ratio of the dose or exposure level of the chemical to a reference dose (RfD) or reference concentration (RfC) established by regulatory agencies or scientific bodies as a safe level of exposure. If the hazard quotient is greater than 1, it suggests that the level of exposure is of potential concern and additional risk assessment may be needed.

(a) The hazard quotient (HQ) is calculated as the daily intake of a chemical divided by its reference dose (RfD). The RfD for 1,1-dichloroethylene is 0.02 mg/kg/day.

The daily intake of 1,1-dichloroethylene can be calculated as:

Intake = concentration × ingestion rate × body weight

Intake = 0.1 μg/L × 2 L/day × 70 kg = 14 μg/day = 0.0002 mg/day

The hazard quotient is:

HQ = Intake / RfD = 0.0002 mg/day / 0.02 mg/kg/day = 0.01

(b) The cancer risk from exposure to 1,1-dichloroethylene can be estimated using the unit risk factor (URF) for this chemical, which is 0.5 per mg/kg/day. The cancer risk is calculated as:

Risk = Intake × URF = 0.0002 mg/day × 0.5 per mg/kg/day = 0.0001

Therefore, the cancer risk is 1 in 10,000.

(c) If the individual drinks this water for 30 years, the total exposure would be:

Exposure = Intake × 365 days/year × 30 years = 2.19 mg

The new hazard quotient is:

HQ = Exposure / (RfD × body weight) = 2.19 mg / (0.02 mg/kg/day × 70 kg) = 1.57

The new cancer risk is:

Risk = Exposure × URF = 2.19 mg × 0.5 per mg/kg/day = 1.10

Therefore, the cancer risk is 1 in 1,000.

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AnswerNegative Partial Charge on: Se, O, O. O-Te is most polar.

Explanation:

O, Se, and Te are all in the same group on the periodic table, which makes this question easier because there is a periodic trend for electronegativity going up and down groups. The more you go down the periodic table, the more shielding affect there is where the electrons closest to the nucleus of an atom block the protons’ positive charge from reaching other electrons. So, O is the most electronegative here, followed by Se, and then Te. So in a bond with O and either of the other two, O will always have a partial negative charge. Se in a bond with Te will also have a partial negative charge, but the difference in electronegativity is biggest between O and Te, and so that is the most polar bond.

According to the electronic configuration and difference in electronegativity, the most polar bond is O-Te.

Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.

Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.

Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.

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Answer:

\(n_{CO(NH_2)_2}=73.3molCO(NH_2)_2\)

Explanation:

Hello.

In this case, given the two balanced chemical reactions:

\(CO(NH_2)_2 (l) \rightarrow HNCO(l) + NH_3(g)\\\\ 6HNCO(l) \rightarrow C_3N_3(NH_2)_3 (l) + 3CO_2(g)\)

We first compute the moles of HNCO from the obtained 1.00 kg of melamine (molar mass 126 g/mol) by considering the 65 % yield:

\(m_{C_3N_3(NH_2)_3}^{theoretical}=\frac{1.00kg}{0.65}=1.54kg\)

\(n_{HNCO}=1.54kgC_3N_3(NH_2)_3*\frac{1000g}{1kg}*\frac{1molC_3N_3(NH_2)_3}{126gC_3N_3(NH_2)_3} *\frac{6molHNCO}{1molC_3N_3(NH_2)_3} \\\\n_{HNCO}=73.3molHNCO\)

Next, we compute the moles of urea in the first chemical reaction:

\(n_{CO(NH_2)_2}=73.3molHNCO*\frac{1molCO(NH_2)_2}{1molHNCO} \\\\n_{CO(NH_2)_2}=73.3molCO(NH_2)_2\)

Best regards.

Answer:

\(\boxed {\boxed {\sf 10.2 \ kJ}}\)

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

\(q=mc \Delta T\)

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

The aluminum block was heated from 23.0 °C to 73.5 °C.

Now we know all three variables and can substitute them into the formula.

\(q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)\)

Multiply the first two numbers. The units of grams cancel.

\(q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)\)

\(q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)\)

\(q= (201.825\ J / \textdegree C)(50.5 \textdegree C)\)

Multiply again. This time, the units of degrees Celsius cancel.

\(q= 201.825 \ J * 50.5\)

\(q= 10192.1625 \ J\)

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\(\frac { 1 \ kJ}{ 1000 \ J}\)

Multiply by the answer we found in Joules.

\(10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}\)

\(10192.1625 * \frac{ 1 \ kJ}{ 1000 }\)

\(\frac {10192. 1625}{1000} \ kJ\)

\(10.1921625 \ kJ\)

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

\(10.2 \ kJ\)

Approximately 10.2 kilojoules of energy would be required.

Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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An average brightness was calculated for each length of graphite tested to get a better understanding of the relationship between the length of the graphite and the brightness of the line it produced.

Graphite has many uses in various industries. Some of its uses include: Pencils, Lubricants, Refractories, Batteries, Electrodes, Nuclear reactors, Aerospace industry.

By calculating the average, it is possible to see if there is a trend in the data and if longer or shorter lengths of graphite produce brighter or duller lines. This information can be useful in determining the best length of graphite to use for a particular task or project.

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Answer:evaluations of seismic data

Explanation:

Answer:

Explanation:

In theory, not much of anything. The vast majority of nitrates are water soluble. Aside, not sure what chemistry level you are at but you will probably be asked to know or memorize some solubility rules. This, for lack of a better phrase, Nitrate rule, is near spot on. With one exception—a rare one—all metal cationic nitrates are soluble in water. All of them. So, assuming you are talking about aqueous, water-based solutions of these salts and mixing them together, I expect nothing to occur. Both solutions, I believe are colorless in water and will thus remain so. If you had say a solution of Iron (III) nitrate and copper (II) nitrate, slightly different story. Both are colorful solutions and I would think you might see blending of colors but no reaction; no precipitate will form. You will probably learn about markers of a chemical reaction. One of these is a color change. Note, you should read this as a change of color from what you previously had. Going from red to blue or colorless to colored (or vice versa) is a strong indication of a reaction (e. g. evidence of bond-breaking and bond-formation). The mere mixing of colors does not constitute a chemical reaction.

The molecules of Li₂O that would produce if you used up 6 atoms of Li(s) is 3.

It is an unstable inorganic and radical compound, used in the manufacture of glass and ceramic.

The balanced equation

\(\rm 6 Li(s) + O_2(g) \rightarrow 3 Li_2O(s)\)

If the atoms of lithium moves up to 6 the lithium oxide molecules will be 3, as you can see in the equation.

Thus, the molecules of Li will be 3.

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Filtration is the process of passing a liquid containing contaminants through a filter to get a cleaner liquid.

After filtration impurities are left on the filter when the liquid goes through it, resulting in a cleaner combination. The procedure of separating one ingredient from another in a compound is used in the purification of chemical compounds or any materials. A mixed sample is turned into fresh samples, each of which contains a single ingredient.Separation procedures are then utilised to separate these compounds. These techniques alter the relative proportions of components in a combination.The type of the compounds and the impurities in them influence the purification of organic chemicals and compounds.

Purification of compounds can be done in other methods as well such as distillation, chromatography, sublimation and crystallisation.

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Full question:

What can you conclude after learning about filtration of solutions of organic compounds? Name other methods of purification of compounds.

Answer:

Look in explanation

Explanation:

I have shown the Lewis diagram as well as the square and sphere representation.

Answer:

International Union of Pure and Applied Chemistry (IUPAC) justified ways how to name organic compounds [ hydrocarbons ].

Oxygen has only a molecular name which is dioxide

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

\(c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M\)

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

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Answer: 9.18 Litres

Standard Temperature and Pressure (STP). Think of this as the perfect environment where the Temp. is 0°C or 273 Kelvin and Pressure is always 1 atm. This is only true in STP.

This question uses the Ideal Gas Equation:

PV=nRT

P= 1 atm

V = ??

T = 273 K (always convert to Kelvin unless told otherwise)

n = 0.410 mol

R = 0.0821 L.atm/mol.K

What R constant to use depends on the units of the other values. (look at the attachments) The units cancel out and only Litres is left. You simply multiply the values.

Answer:

9.18 liters .............

heya

Answer:

C

Oxygen gas is limiting.

C(s) + O

2

→CO

2

(g)

No. of moles of carbon =

12

36

=3 moles

No. of moles of oxygen =

32

32

=1 moles

So, 2 moles of carbon is left and oxygen will be completed.

So, O

2

is limiting reagent.

Answer:

14.5

Explanation:

not sure how I got it but I hope this helped!

The volume of the rectangle has been given by the product of length, breadth, and height. The mass of the air with a density of 1180 g/m³ is 708.57 gms.

The density of the substance has been known to be given by the mass's ratio to the volume. The formula for density is given as,

Density = Mass ÷ Volume

Given,

Density = 1180 g/m³

The volume of the rectangle = L×B×H

Substituting the values above,

Mass = Density × Volume

M = 1180 × 1.2 × 10⁻² × 8.34 × 10¹ × 6.0 × 10⁻¹

= 708.57 gm

Therefore, the mass of the air is 708.57 gms.

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Answer:

Explanation:

c

option a,b,c are correct.gather your belongings,turn off all hot plates & other heating apparatus, turn off all water these are options that must be done in case of a fire alarm or other evacuation .

To guarantee your safety, take prompt action when the fire alarm goes off. The fire alarm system was created and constructed to give you a head start so you may safely leave the building in an emergency.Never dismiss a warning or believe it to be fake or the outcome of a test.The nearest and safest exit and/or stairs must be used by everyone to leave the building.Never leave during a fire alarm activation in an elevator.Once outside, make your way away from the structure. Gather on the sidewalk of the building next door or across the street.The firefighters and fire engines will be working in front of the structure. Don't stand in their way as they enter the building.

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The balanced equation for the breakdown of solid ammonium hydroxide into gaseous ammonia and liquid water is:

\(NH_{4} OH\)(s) → \(NH_{3}\) (g) + \(H_{2}O\) (l)

This equation indicates that one molecule of solid ammonium hydroxide (\(NH_{4} OH\)) decomposes to form one molecule of gaseous ammonia (\(NH_{3}\)) and one molecule of liquid water (\(H_{2}O\)).

The type of reaction that is occurring is a decomposition reaction. This is because one compound, ammonium hydroxide, is breaking down into two simpler substances, ammonia and water. Decomposition reactions can be induced by heat, light, or electricity and are a common type of chemical reaction in nature. In this case, the breakdown of ammonium hydroxide is likely to be endothermic since heat is required to break the bonds within the compound.

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pKa value =4.76Ka

=1.73×10 −5 !!    PH=4−PKa=14−4.76=9.24 for a weak monoprotic acid is titrated with 0.100 M NaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. After 25.0 mL of the base is added, the pH of the solution is 3.52.

One way to describe an acid's potency is by its pKa value. The acid dissociation constant, or pKa, is the negative log of the pKa value. A stronger acid is indicated by a lower pKa value. In other words, a lower value means that the acid dissociates in water more completely.

The logarithm of Ka's negative value is denoted as pKa. The logarithmic inverse of the H+ concentration is pH. Acidity indication. An acid's pKa value tells you if it's a strong acid or a weak acid. The pH scale shows how acidic or alkaline a system is.

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The equation C2H4(g) + H2(g) + C2H6(g) → Caco3(s) + Cao(s) + CO2(g) shows an increase in entropy due to the formation of a gas as a product. Option A

In this equation, the reactants on the left-hand side consist of gases (C2H4 and H2), while the products on the right-hand side include a solid (Caco3) and a gas (CO2).

When a reaction involves a change from gaseous to solid or liquid states, there is typically a decrease in entropy because the particles become more ordered and constrained in the solid or liquid phase.

Conversely, when a reaction involves the formation of gases, there is generally an increase in entropy because gases have higher degrees of molecular motion and greater freedom of movement compared to solids or liquids.

In the given equation, the reactants include three gaseous compounds (C2H4, H2, and C2H6), and one of the products is a gas (CO2). Therefore, the overall entropy of the system increases during this reaction.

The equation Fe(s) + S(s) → FeS(s) does not show an increase in entropy. Both the reactants (Fe and S) and the product (FeS) are solids. Since solids have lower entropy compared to gases or liquids, the entropy of the system does not increase in this reaction. Option A

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Answer:

The possible source of errors as regard the Procedure are:

Explanation:

The procedure referred to should a experiment with regards heat

The possible source of errors as regards the Procedure are:

Answer:

CO2 AND H20 I THINK IM 99 PERCENT SURE