Give the IUPAC name for the following structure

Answer:

[tex]M=0.0637M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]

Finally, the resulting molarity in 30.8 mL (0.0308 L):

[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]

Regards.

Answer:

H2 consumed 4.22 mol

N2 produced 59.107 g

Explanation:

Balanced equation:

2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)

H2O MW = 18.02 g/mol

N2 MW = 28.01 g/mol

To determine the moles of H2O produced, the following formula should be used:

[tex]MW=\frac{mass}{mol}[/tex]

The value of moles is cleared:

[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]

2 mol H2      ⇒ 1 mol N2

4.22 mol H2 ⇒ X

[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]

To calculate the mass of H2 consumed, the molecular weight equation is used again:

[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]

Answer:

f = 33.34 Hz

Explanation:

A wave has a period of 0.03 seconds. It is required to find the frequency of a wave. The relation between time period and frequency is inverse. The time period of a wave is given by :

T = 1/f, f = frequency of wave

[tex]f=\dfrac{1}{T}\\\\f=\dfrac{1}{0.03}\\\\f=33.34\ Hz[/tex]

So, the frequency of the wave is 33.34 Hz.

Answer:

The correct answers are 4.93 %, 4.72 % and 4.93 %.

Explanation:

Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,  

% error = (actual value - calculated value) / actual value * 100

By 1st student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41 - 15.12) / 14.41 * 100  

= 0.71/14.41 * 100

= 4.93 %

By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,  

% error = (14.41-15.09)/14.41 * 100

= 0.68/14.41 * 100

= 4.72 %

By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41-15.12)/14.41 * 100

= 0.71/14.41 * 100

= 4.93 %

Answer:

True

Explanation:

Due to the arrangement of the molecule, a carbon dioxide molecule is non-polar.

Answer:

C = 98%

Explanation:

Hello,

To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.

Mass of salt recovered = 1.47g

Initial mass of salt = 1.50g

Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100

Percentage of salt recovered = (1.47 / 1.50) × 100

Percentage of salt recovered = 0.98 × 100

Percentage of salt recovered = 98%

The percentage of salt recovered is equal to 98%

Answer:

The answer is Hg.

Explanation:

Symbol for Mercury is Hg.

Answer:

below

Explanation:

28th 10;24 am

If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later at 12A.M.

This movement is from the Moon's orbit, which takes 27 days, 7 hours and 43 minutes to go full circle. It causes the Moon to move 12–13 degrees east every day. This shift means Earth has to rotate a little longer to bring the Moon into view, which is why moonrise is about 50 minutes later each day.

So knowing that moonrise is about 50 minutes later each day, we have:

[tex]7+50 minutes = 7:50\\8:40\\9:30\\10:20\\11:10\\12:00 A.M[/tex]

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Answer:

A. Solubility of calcium sulfite increases

B. Solubility of calcium fluoride increases

C. Solubility of Silver bromide decreases

Explanation:

The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.

Case 1. Calcium sulfite

The dissociation reaction of CaSO3 is given below:

CaSO3 ----> Ca²+ + SO3²-

SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;

H+ + SO3²- ---> HSO3-

This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.

Case 2. Calcium fluoride

The dissociation reaction of calcium fluoride (CaF2) is shown below.

CaF2 ----> Ca²+ + 2F-

Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:

H+ + F- ---->. HF

Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.

Case 3: Silver bromide

The dissociation reaction of AgBr is as follows:

AgBr ----> Ag+ + Br-

The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.

The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.

Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.

Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.

However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.

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Answer:

Chemical Change

Explanation:

Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.

Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Explanation:

[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]

Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]

The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

[tex]K_a[/tex] for  [tex]CH_3COOH[/tex] :

[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]

[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]

[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

find the given attachment

Answer:

6.91 × 10¹⁴ s⁻¹

Explanation:

Step 1: Given data

Wavelength of the radiation absorbed by the cone (λ): 434 nm

Step 2: Convert the wavelength to meters

We will use the relationship 1 m = 10⁹ nm.

[tex]434nm \times \frac{1m}{10^{9}nm } =4.34 \times 10^{-7} m[/tex]

Step 3: Calculate the frequency (ν) of the radiation

We will use the following expression.

[tex]c = \lambda \times \nu[/tex]

where,

c is the speed of light (3.00 × 10⁸ m/s)

[tex]c = \lambda \times \nu\\\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}m/s }{4.34 \times 10^{-7}m }= 6.91 \times 10^{14} s^{-1}[/tex]

Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness

just like it does with chemical phenomena

=)

Answer:

Part A

The natural abundance of Si-30 = 3.1

Part B

The mass of Si-30 = 29.9551 amu

Explanation:

Part A

The sum of the natural abundances = 100

Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.

Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%

Si-30's mass and natural abundance are unknown.

Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.

Part B

The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.

Molar mass of Silicon normally = 28.0855 amu

Let the mass of Si-30 be m

28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)

28.0855 = 27.14790435 + 0.0313m

0.0313m = 28.0855 - 27.14790435 = 0.93759565

m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu

Hope this Helps!!

Part A:  The natural abundance of Si-30 = 3.1

  Lets solve the question:

Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.

Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.

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Answer:

Plasmas and solids are both made up of cation-anion pairs. Solids and plasmas are both made up of electrons and cations. Solids are made up of cation-anion pairs, but plasmas are not.

Answer:

Solids are made up of cation-anion pairs, but plasmas are not.

Explanation:

Answer:

D

Explanation:

Color, texture, and density are all physical properties but reactivity is a chemical property so the answer is D.

Answer:

THE STANDARD HEAT OF NEUTRALIZATION OF THE BASE SODIUM HYDROXIDE BY THE ACID HYDROGEN TRIOXONITRATE V ACID IS -56 kJ / mol.

Explanation:

Volume of 0.3 M NaOh = 100 mL

Volume of 0.3 M HNO3 = 100 mL

Initail temp of NaOH and HNO3 = 35 °C = 35 + 273 K = 308 K

Final temp. of mixture = 37 °C = 37 + 273 K = 310 K

We can make the following assumptions form the question given:

1. specific heat of the reaction mixture is the same as the specific heat of water = 4.2 J/g K

2. the toal mass of the reaction mixture is 200 mL = 200 g since no heat is lost to the calorimeter or surrounding.

3. initail temperature of the reaction mixture is equal to the average temperature of the two reactant solutions

= ( 308 + 308 /2) = 308 K

4. Rise in temeperature for the reaction = 310 -308 K = 2 K

Then the total heat evolved during the reaction = mass * specifc heat capacity * temperature  change

Heat = 200 g * 4.2 J/g K * 2 K

Heat = 1680 J

EQUATION FOR THE REACTION

HNO3 + NaOH -------> NaNO3 + H20

From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to prouce  mole of water.

100 mL of 0.5 M HNO3 contains 100 * 0.3 /1000 = 0.03 mole of acid

This result is same for the base NaOH = 0.03 mole of base

So therefore,

0.03 mole of acid will react with 0.03 mole of base to produce 0.03 mole of water to evolved 1680 J of heat energy.

The production of 1 mole of water will evolve 1680 / 0.03 J of heat

= 56 000 J or 56 kJ of heat energy per mole of water.

So therefore, 1the standard heat of neutralization of sodium hydroxide by trioxoxnitrate V acid is -56 kJ/mol.

Answer:

Explanation:

Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.

SOo, when vinylcyclohexane is treated with  NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical  with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .

This   stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.

The diagram emphasizing more on the above explanation can be seen in the attached image below

Answer:

Ice (frozen water) is part of the hydrosphere, but it's given its own name, the cryosphere.

The cryosphere is part of the hydrosphere of the Earth system. The correct option is D.

The cryosphere contains all the frozen parts of the earth. The term is made up of the Greek word “krios” which means cold. All the frozen water of the oceans and snow comes under the cryosphere.

The atmosphere contains all spheres, it is an envelope of gases. The geosphere is the land part of the earth, and the biosphere is the part where the living part is present.

The cryosphere is h habitat of many living creatures, and the climate of the earth is highly dependent on this sphere. The warmth of the earth is increasing and the cryosphere part is decreasing day by day, which is having problems for many animals.

Thus, the correct option is D, hydrosphere.

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Answer:

0.067M Fe(NO3)3

Explanation:

A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.

The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.

The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3

That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:

0.200M / 3 =

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

0.263M of CH₃COOH is the concentration of the solution

Answer:

218.3 g/mol

Explanation:

Boiling point elevation occurs when a solute is added to a solvent increasing the boiling point of the solution with regard to the pure solvent.

The law is:

ΔT = Kb×m×i

Where ΔT is change in temperature (3.67°C), Kb is the boiling point constant of the solvent (4.95°C/m), m is molality of the solution and i is Van't Hoff factor (1 for a non-electrolyte).

3.67°C = 4.95°C/m×m×i

0.7414m = molality of the solution (Moles solute / kg solvent).

As the mass of the solvent is 50.0g = 0.0500kg:

0.7414m = Moles solute / 0.0500kg

0.0371 = moles of solute

As the mass of the solute is 8.1g, molar mass of the solute (Ratio between mass in g and moles) is:

8.1g / 0.0371mol =

Answer:

It is the portion of internal energy that can be transferred from one substance to another.

Thermal energy is the portion of internal energy that can be transferred from one substance to another.

Thermal energy is the energy an object posses which is as a result of particles movement within it.

It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.

Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.

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Answer:

The information has been changed to support a claim

Explanation:

Answer:

endothermic

Explanation:

Heat is added to make the process possible.

Answer:

An early theory describing properties of atoms.

Explanation:

Apex