Each individual candy/isotope is the same and will take the same amount of time to decay.
a. true
b. false

Student e accounted for the equivalent weight found for succinic acid by analyzing its titration with [tex]NaOH[/tex] (aq) and concluding that it is diprotic.

One mole of succinic acid required two mole of base to neutralize completely. In this titration phenolphthalein indicator is used to observe the completion of the reaction. In an aqueous solution, succinic acid readily ionize to form succenate. Succinic acid undergoes two successive deprotonation reaction.

The succinic acid results in succinate ( dicarboxylic acid dianion ) which is form by removal of a proton from both carboxy groups of succinic acid.

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[tex]NO_{2}[/tex](g) + OH(g) = [tex]HNO_{3}[/tex] (aq) (nitric acid)

Given,

Pressure of [tex]NO_{2}[/tex] = 4.5x10-8 atm,

volume = 1.3x109 L, and

temp = 10ºC + 273 = 283K

moles. NO2 =? = n

PV = nRT

n = PV/RT = (4.5x10-8 atm)(1.3x109 L) / (0.0821 Latm/Kmol) (210K)

n = 3.4x10-18 moles NO2 = mols HNO3

mass HNO3 = 3.4 x 10 18 moles x 31 g/mol = 1.05 x 10 16 g.

The inorganic substance nitric acid has the formula HNO3. This mineral acid is extremely corrosive. [5] Even though the chemical is colorless, older samples often have a yellow tinge from breakdown into nitrogen oxides. Nitric acid is 68 percent concentrated in water in the majority of commercially available products. Fumigating nitric acid is the term used to describe a solution that contains more than 86 percent HNO3. Depending on the quantity of nitrogen dioxide, fuming nitric acid is further classified as red fuming nitric acid at concentrations above 86 percent or white fuming nitric acid at concentrations above 95 percent. The main chemical used in nitration, or adding a nitro group usually to an organic molecule, is nitric acid.

How many grams of nitric acid are produced from this reaction if it reduces the partial pressure of no2 from a 1. 3×109l volume of air over a city by 4. 5×10−8atm when the temperature is 10∘c?

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The sample of gas that is unreactive to flame tests and chemically reactive with limewater and bromothymol blue is Carbon dioxide.

Flame tests are performed to identify the unknown metals or metalloids.  Upon burning on the flame, a compound gives characteristic colour, that helps in identification.

Carbon dioxide extinguishes the flame itself when placed upon it. And it is known to turn the lime water milky when reacts. Carbon dioxide also changes the colour of bromothymol blue. The change in colour is observed due to certain chemical changes like change in pH, formation of carbonic acid and conversion into more acidic form.

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Answer:

four

Explanation:

i said so

Answer:

synergy

Explanation:

when two substances combine effects to be greater than just their sum of their individual effects

Taking into account the reaction stoichiometry and the definition of STP, 72.8 L of CO₂ gas is produced once the gas cools to STP.

In first place, the balanced reaction is:

CaCO₃  → CaO + CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CaCO₃ form 1 mole of CO₂, 3.25 moles of CaCO₃ form how many moles of CO₂?

[tex]moles of CO_{2} =\frac{3.25 moles of CaCO_{3}x 1 mole of CO_{2}}{1 moleof CaCO_{3}}[/tex]

moles of CO₂= 3.25 moles

Then, 3.25 moles of CO₂ are formed when 3.25 moles of CaCO₃ decomposes.

Now, you can apply the following rule of three: if by definition of STP conditions 1 mole of CO₂ occupies a volume of 22.4 liters, 3.25 moles occupies how much volume?

[tex]volume= \frac{3.25 molesx22.4 L}{1 mole}[/tex]

volume= 72.8 L

Finally, 72.8 L of CO₂ gas is produced once the gas cools to STP.

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The acidic functional group that can dissociate and release h into a solution is -COOH.

From the characteristics of -COOH group we can conclude it as the acidic functional group that can dissociate and release H into a solution.

Disclaimer: The question was incomplete

Which is an acidic functional group that can dissociate and release h into a solution?

A. -OH

B. -COOH

C. -NH2

D. -SH

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The latent heat of vaporization 5gram of steam converted to liquid at 100°C is 11.3 KJ.

The latent heat of vaporization for a given substance tells you how much energy is required for one mole of that substance to undergo a phase transition or go from a liquid to a gas, at its boiling point.

Joules per gram, an alternative to the more popular kilojoules per mole, are used to express the latent heat of vaporization for water.

Therefore, we must determine how many kilojoules per gram are needed for a certain sample of water to transition from a liquid to a vapor at its boiling point.

As you know, the conversion factor that exists between Joules and kilojoules is 1 kJ = 10³ J

2260 J/g will be equivalent to

                               [tex]2260 \frac{J}{g} . \frac{1kJ}{1000J } = 2.26 kJ/g\\\\[/tex]

As we know,

2260 = 2.26 . 10³

which means that 2.26 .10³ = 2260J

This is the latent heat of vaporization 5gram of water= 2260J/g × 5g

                                                                                   = 11,300J

                                                                                   = 11.3 KJ

Therefore, the latent heat of vaporization 5gram of steam converted to liquid at 100°C is 11.3 KJ.

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The pH of a 1. 0 l buffer prepared by adding 0. 100 moles of NaOH to 0. 250 moles of HF. ka for HF = 3. 5 × 10−4 is 3.86.

The Henderson hasselbalch equation can be expressed as

pH = pKa + log [base]/ [acid]

Firstly we will calculate the value of pKa

pKa = -logKa

Given,

Ka = 3. 5 × 10−4)

pKa = -log(3. 5 × 10−4)

pKa = 3.46

Now, we will calculate the value of log [base]/ [acid]

Given,

[base] = 0.10m

[acid] = 0.25m

log [base]/ [acid] = log(0.10/0.25)

= 0.4

Putting the values in handerson hasselbalch equation,

pH = 3.46+0.4

= 3.86

Thus, we find that the value of pH of a 1. 0 l buffer prepared by adding 0. 100 moles of NaOH to 0. 250 moles of HF is 3.86.

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When 1. 0 l of 0. 00010 m NaOH and 1. 0 l of 0. 0014 m mgso4 are mixed, there will be no precipitate formed.

The precipitate is the solid concentration of a substance that is collected over a solution.

First, we determine the concentration of magnesium and hydroxide

(Mg2+) = 7.00 × 10⁻⁴

(OH−) = 5.00 × 10⁻⁵

Now, we calculate the solubility quotient

Qc = (Mg2+) (OH−) ²

Qc = 7.00 × 10⁻⁴ x (5.00 × 10⁻⁵)²

Qc = 1.75 x 10⁻¹²

The solubility product of the magnesium hydroxide is 1.80 x 10⁻¹¹ which is more than the solubility quotient. Thus, there will be no precipitate form.

Thus, there will be no precipitate formed because the solubility quotient we calculated is less than the solubility product.

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The value of ''n" in the hydrate formula is 10

Mass of anhydrous NA2B4O7 = 2.64g

Molar mass of NA2B4O7 = 201.224 g/mol

Number of Moles of anhydrous NA2B4O7 = Mass of NA2B4O7 / Molar mass of NA2B4O7

                                                     = 2.64 / 201.224

                                                     = 0.0131 mol

Mass of water in 5.00g of NA2B4O7 = 5.00 - 2.64

                                                              = 2.36 g

Moles of water = 2.36 / 18.015

                          = 0.1307 mol

The whole number ratio = moles of water / moles of anhydrous sodium tetra borate

                                           = 0.1307 / 0.0131

The whole number ratio = 10

Hence the value of n in the hydrate formula is 10.

NA2B4O7. 10H2O

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Answer:

82.4 s

Explanation:

Find the NUMBEr of half lives...then multiply by 54.3

2.27 = 6.5 (1/2)^n

log (2.27/6.5) / log (1/2) = n = 1.52 half lives

1.52 * 54.3 = 82.4 s

Rapid breathing pays back the oxygen debt by breaking down lactic acid. creatine glycogen oxygen glucose.

Even after exercise must be finished, the "oxygen debt" could be paid when laborious breathing and an elevated heart rate were needed to eliminate lactic acid and replenish depleted energy stores.

The oxygen debt would be the quantity of oxygen needed to replenish the body's oxygen stores and remove the lactic acid. It can take anywhere between a few hours for little activity and several days after just a marathon when an individual has been exercising to start repaying an oxygen debt.

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Succinate dehydrogenase is the citric acid cycle considered part of aerobic metabolism even though oxygen does not appear.

The respiratory complex II, also known as succinate dehydrogenase (SDH), succinate-coenzyme Q reductase (SQR), or SDH, is an enzyme complex that is present in the inner mitochondrial membrane of eukaryotic and numerous bacterial cells. SDH converts succinate to fumarate as part of the citric acid cycle. SDH shares structural similarities with fumarate reductase, an enzyme that catalyzes the reverse process during anaerobic respiration in bacteria. (1997, Hagerhall). Succinic semialdehyde dehydrogenase impairment is brought on by ALDH5A1 gene mutations. Instructions for creating the succinic semialdehyde dehydrogenase enzyme are found in the ALDH5A1 gene.

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Nuclear decay is a radioactive decay where energy is released by unstable nuclei. After thorium undergoes decay, actinium-228 will be the final product. Thus, option a is correct.

Alpha radioactive decay is the loss of two protons and neutrons from the element that results in a decrease of the atomic mass by 4 and atomic number of the element by two.

²³² Th₉₀ → ²²⁸Ra₈₈ + ⁴α₂

Beta radioactive decay is the increase of the atomic number but not the atomic mass of the element.

²²⁸Ra₈₈ → ²²⁸Ac⁸⁹ + β⁻

Therefore, option a. actinium-228 is the final product of the decay.

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Your question is incomplete, but most probably your full question was, In the initial sequence of thorium-232 decay, an alpha particle is emitted followed by a beta particle. What is the product of these two decay steps?

²³² Th → α particle + β particle + ?

Answer:

29 mL

Explanation:

The question needs us to find the volume of the liquid. The equation for volume using density and mass is:

Volume = Mass / Density

We can substitute the given values for density and mass into the equation:

[tex]V=\frac{25\ g}{0.87\ g/ml}[/tex]

[tex]V\approx29\ mL[/tex]

The answer we obtained (29 mL) is rounded to two significant figures. When multiplying or dividing, the amount of significant figures in the final answer is always the least amount of significant figures in one of the values.

Below are the significant figure rules:

Nonzero digits will always be significant (eg. 54 --> 2 significant figures)

Zeroes at the beginning of a number will never be significant (eg. 0.1 --> 1 significant figure)

Zeroes between two nonzero digits will always be significant (eg. 504 --> 3 significant figures)

Zeroes following a number will always be significant if the number contains a decimal point (eg. 40.0 --> 3 significant figures)

In the polymerization reaction, the lone pair electrons on the NH₂ groups of hexanediamine attack the C=O groups of the dicarboxylic acid in a nucleophilic substitution reaction as shown in the image.

Hydroxide is added to remove any H⁺ ions present and keep the hexanediamine in the deprotonated form, so that the NH₂ lone pair electrons are available for reaction.

If hydroxide is not added, the NH₂ groups will get protonated by H⁺  ions present to give NH₃⁺ groups, which cannot react.

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option (c) emulsion is the right answer

Emulsion is a classification of milk.

In physical chemistry, an emulsion is a combination of two or more liquids in which one of the liquids is present as microscopic or ultramicroscopic droplets dispersed throughout the other.

Emulsions form as a result of the cleansing action of soaps.

(ii) The emulsification process is how lipids are broken down in the intestines.

(iii) Disinfectants and antiseptics combine with water to generate emulsions.

(iv) The emulsification technique is utilized to create medications.

When two immiscible liquids, such as oil and water, are stirred together with an emulsifier—which might be a protein, phospholipid, or even a nanoparticle—emulsion is created.

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Taking into account the definition of calorimetry, the specific heat of the metal is 0.14 [tex]\frac{J}{gC}[/tex].

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

Q is the heat exchanged by a body of mass m.

c is specific heat substance.

ΔT is the temperature variation.

In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also decreases; and the same happens when either of the two decreases .).

The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body.

Specific heat of the metal

In this case, you know:

Replacing in the expression to calculate heat exchanges:

175 J= c× 50× 25 C

Solving:

c=[tex]\frac{175 J}{50 gx25C}[/tex]

c=0.14 [tex]\frac{J}{gC}[/tex]

Finally, the specific heat of the metal is 0.14 [tex]\frac{J}{gC}[/tex].

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If we takes 11. 2 kj of energy to raise the temperature of 145 g of benzene from 23. 0°c to 68. 0°c, then the specific heat of benzene is

1716.475 J/kg.°C.

Specific Heat capacity is given as

Q = cm(t₂-t₁) ................[tex]equation 1[/tex]

Where Q = quantity of heat, c = specific heat of benzene, m = mass of benzene, t₁  = initial temperature, t₂ = final temperature.

[tex]C[/tex] = [tex]\frac{Q}{m_{(t2-t1)} } ...........equation 2[/tex]

Given:

[tex]Q = 11.2 kJ = 11200 J \\\\m= 145\\\\g = 0.145 kg\\\\t_{2} = 68.0 °C\\\\t_{1} = 23.0 °C[/tex]

Substitute into equation 2.

[tex]C = \frac{11200}{[0.145(68-23)]}\\\\ C= \frac{11200}{6.525}\\\\ C= 1716.475 J/kg °C[/tex]

Hence the specific heat of benzene = [tex]1716.475 J/kg °C[/tex]

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The energy required to melt 20 g of water (ice) starting at 0° C is 6680 J.

Heat is a form of energy that is possessed by a system by virtue of its temperature. Its unit is Joule.

Enthalpy is the amount of the heat content of a system.

It is the amount of heat required to increase the temperature of a system by 1° C. It is an extensive property. Its unit is J/°C

Heat Capacity is given by

[tex]c= \frac{q}{\Delta T}[/tex]

where q = heat required

[tex]\Delta T[/tex] = change in temperature

Enthalpy of Fusion is the enthalpy change when 1 mole of a solid substance changes into its liquid state at its melting point.

Here, water in solid form (ice) melts into liquid form at 0°C

[tex]H_{2} O (s) \rightarrow H_{2} O(l) \\\Delta H_{fus } =334Jg^-^1[/tex]

We know,

[tex]\Delta H = \frac{m}{q}\\[/tex]

where m =  given mass

q = heat required

[tex]q = \Delta H \times m[/tex]

q = 334 [tex]Jg^-^1[/tex] x 20 g

    = 6680 J

Thus, the heat required to melt 20 g of water at 0°C is 6680 J

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pH value of a 0. 011 m naf solution is 8.57.

Solution

This problem uses the relationship between Kb and the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:

Kb = KaKb

1.00 x 10^-14 = 7.2 x 10^-4(x)

x = 1.39 x 10^-11

We now need to calculate the [OH¯] using the Kb expression:

1.39 x 10^-11 = x^2 / (0.30 - x)

The denominator can be neglected. Thus, x is 3.73 x 10^-6.

pOH = -log 3.73 x 10^-6 = 5.43

pH = 14-5.43 = 8.57

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Answer:

1.1 x 10⁴ J

Explanation:

To calculate eth energy needed, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy/heat (J)

-----> m = mass (g)

-----> c = specific heat (4.184 J/g°C)

-----> ΔT = change in temperature (°C)

You can plug the given values into the equation and solve.

Q = mcΔT

Q = (75.0 g)(4.184 J/g°C)(55.0 °C - 20.0 °C)

Q = (75.0 g)(4.184 J/g°C)(35.0)

Q = 11,000 J

Q = 1.1 x 10⁴ J

A monosaccharide that contains 6 carbon atoms, one of which is in an aldehyde group, is classified an Aldohexose.

Monosaccharides of specific sizes may be indicated by names composed of a stem denoting the number of carbon atoms and the suffix -ose.

For example, the terms triose, pentose, and hexose signify monosaccharides with, respectively, three, four, five, and six carbon atoms. Monosaccharides are also classified as aldoses or ketoses.

Those monosaccharides that contain an aldehyde functional group are called aldoses; those containing a ketone functional group on the second carbon atom are ketoses. Combining these classification systems gives general names that indicate both the type of carbonyl group and the number of carbon atoms in a molecule.

. Glucose and fructose are specific examples of an aldohexose and a ketohexose, respectively.

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The concentration of [tex]NaCl[/tex] solute particles in water will provide an isotonic eyedrop solution is 0.9% w/v

The isotonic solution is calculated by multiply the quantity of each drug in the prescription by its sodium chloride equivalent E , and subtract this value from the concentration of sodium chloride which is isotonic with body fluids . For a solution to be termed isotonic ( equal tone ) it must have the same osmotic pressure as a specific bodily fluid.

Example : The concentration of [tex]NaCl[/tex] solute particles in water will provide an isotonic eyedrop solution is 0.9% w/v

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The two ways were,

A cathode-ray tube, often known as a CRT, is a vacuum tube that houses one or more electron cannons that produce electron beams that are then bent to produce pictures on a phosphorescent screen. The visuals might be pictures (from a television or computer display), electrical waveforms (from an oscilloscope), radar targets, or other occurrences.

Tomson came to the conclusion that atoms can be divided into their constituent corpuscles (atoms are made up of smaller particles).

The electron was discovered in 1897 by J. J. Thomson.

His "plum pudding" concept from 1904 postulated that the positive charge contains embedded electrons.

With this model, he disproved his first theory (the atom was composed of immaterial vortices).

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Explanation:

harmful effects of ozone layer depletion on human being are as follows :

i. It increases in certain type of cancer.

ii. It increases in eyes cataracts.

iii. It increases in immune deficiency disorder.

iv. It changes the climate.

The reaction of 75.0 g p4 with excess chlorine gas produces 110 g pcl3 in lab. The percent yield for the reaction is 32.7%.

P4+ 6 Cl2 — 4PCl3

moles of P4 in 75 g

n=m/M

where n = moles; m= mass; M = molar mass

M(P4) = 4× 30.974 g/ mol

P = 123.896g/mol

n can be given as

n(P4) = (75) / 123.896

= 0.60535 mol

Total=

0.60535 mol P4 × 4 mol/1 mol P4

= 2.4214 mol PCl3

mass of PCl3

m= n×M

M = 30.974 g/mol P+(3× 35.45g/mol Cl)

= 137.32g/mol

m= 2.4212 mol× 137.32g/mol

=333 g

The yield of PCl3 is 333 g.

The theoretical yield of PCl3 is 110 g.

percent yield can be given as

(percent yield/ theoretical yield) × 100

%yield = (333/110) ×100

Thus, the percent yield of PCl3 is 32.7%.

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Answer: a. 65.37 mL
b. 5.3 cm
c. 2.08 km (plus) or 1.64 km (minus)
d. 18°C (plus) or 16°C (minus)

Explanation:

The way significant figures work when you're adding or subtracting, is you go to the answer with the least amount of decimal places. Take a. for example; you have 65.316 (3 places after the decimal) and 0.05 (2 places after the decimal place). Based off of this, you would round your answer to two places after the decimal. And when you have a whole number, like in d. you round to the nearest whole number, since there aren't any decimal places to round your answer to.