1. What is the frequency of light with a wavelength of 1064 nm?
2. How many photons of 1064 nm light are contained in a 535 Joules of this light?

Answer:

C. 100

Explanation:

Biochemical researches and studies have found out that an average health hotdog has a calorie of between 100 and 150 which is usually dependent on the additives.

Since the nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories then it truly contains such amount of calories. The standard number of calories present in a hotdog is independent of the amount eaten by individuals.

Explanation:

We have to put out the balanced chemical equation before proceeding. The equation for the dissociation of sodium sulfide in water is given as;

Na₂S  --> 2Na⁺ + S²⁻

From the stoichiometry of the reaction we can tell that;

1 mol of Na₂S produces 2 moles of Na⁺ and 1 mol of S²⁻.

Sodium ion:

1 mol of Na₂S  = 2 mol of Na⁺

2,05 = x

upon solving for x;

x =  2 * 2.05 = 4.10 moles

Sulfide ion:

1 mol of Na₂S  = 1 mol of S²⁻

2.05 = x

upon solving for x;

x = 1 * 2.05 = 2.05 moles

Answer:

HEY ITS RIGHT ABOVE!!!

Explanation:

i did 1.85 instead of 2.05 and it was right... im so sorry

Answer:

Face-centered Cubic Unit Cell (FCC) is consists of 8 atoms at the corners (one atom at each corner) and 6 atoms are present at the face-center (one atom at each face). Each eight atoms at corners contribute 1/8 and six atoms at faces contribute  1/2 in each unit cell.

Answer:

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

Explanation:

The general schemefor a reaction is given as;

Reactants --> Products

In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.

The equation is given as;

AgNO3 + Na2SO4 --> Ag2SO4

The other poduct formed in this reaction is NaNO3.

The full reaction is given as;

AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3

The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.

The balanced equation is given as;

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

In this equation, there are equal number of moles of the atoms on both sides.

Answer:

C

Explanation:

Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.

The molecular equation is :[tex]HI(aq)+KOH(aq) -- > KI (aq) + H_2O (l)[/tex]

The net ionic equation is: [tex]H^+(aq) + OH^-(aq) -- > H_2O(l)[/tex]

When hydroiodic acid reacts with potassium hydroxide, this will result in the formation of a salt i.e. potassium iodide, and water is obtained as a by-product.

The molecular equation can be represented as:

[tex]HI(aq)+KOH(aq) -- > KI (aq) + H_2O (l)[/tex]

The net-ionic equation can be represented as:

[tex]H^+(aq) + OH^-(aq) -- > H_2O(l)[/tex]

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A molecular equation is a balanced chemical equation that represents a chemical reaction by showing the complete chemical formulas of all reactants and products involved. The net ionic equation for the reaction is as follows:

[tex]H+(aq) + OH-(aq) = H_2O(l)[/tex]

The molecular equation for the reaction between hydroiodic acid (HI) and potassium hydroxide (KOH) can be written as follows:

[tex]HI(aq) + KOH(aq) = KI(aq) + H_2O(l)[/tex]

In this equation, (aq) represents aqueous solutions, indicating that the substances are dissolved in water, and (l) represents a liquid phase for water.

To write the net ionic equation, we need to remove the spectator ions that do not participate in the actual chemical reaction. In this case, potassium ion (K+) and iodide ion (I-) are spectator ions, meaning they appear on both sides of the equation without undergoing any change.

Therefore, the net ionic equation for the reaction is as follows:

[tex]H+(aq) + OH-(aq) = H_2O(l)[/tex]

In the net ionic equation, H+ represents the hydrogen ion from hydroiodic acid, and OH- represents the hydroxide ion from potassium hydroxide. These ions combine to form water ([tex]H_2O[/tex]) as the only product.

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Answer:

Partial pressure (Ar) = 316.1mmHg

Explanation:

In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.

Moles of gases:

PV = nRT

P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm

V = 1.66L

n = Moles gases

R = 0.082atmL/molK

T = 54.9°C + 273.15K = 328.05K

PV = nRT

1.1974atm*1.66L = n*0.082atmL/molK*328.05K

Knowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:

1.13g = 2.016X + 39.948Y (1)

Where X = moles of hydrogen and Y = moles of Argon.

Also we can write:

0.0739moles = X + Y (2)

Total moles of the sample are moles of hydrogen + moles Argon

Replacing 2 in 1:

1.13g = 2.016(0.0739-Y) + 39.948Y

1.13 = 0.1564 - 2.016Y + 39.948Y

0.9736 = 37.932Y

0.02567 = Y = moles of Argon

As total moles are 0.0739moles, mole fraction of Ar in the sample are:

XAr = 0.02567mol / 0.0739mol

X Ar = 0.347

Last, partial pressure of Ar = X Ar * total pressure.

Partial pressure (Ar) = 0.347*910.0mmHg

Answer:

work is done on the system by the surroundings

Explanation:

An endothermic process is any process in which energy is absorbed by the system from the surrounding, usually in the form of heat energy. Energy intake is usually associated with bond braking, and the energy from the surrounding goes into breaking the chemical bond between atoms int the reaction. The quantity of heat absorbed in any endothermic process can be calculated for by analyzing the Gibbs free energy of the reaction system. Endothermic process is mostly associated with heat energy intake by the system from the surrounding, but can also be used to refer to any system that undergoes any form of energy input into the system, from the surrounding even in the form of work.

Answer:

a

Explanation:

its made up of carbon and hydrogen

Answer:

Rate law: [tex]r=k[A]^2[/tex]

Integrated rate law: [tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]

Rate constant: [tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]

Explanation:

Hello,

In this case, since the slope is obtained by plotting 1/[A] and it has the units L/(mol*s) or 1/(M*s), we can infer the reaction is second-order, therefore, its rate law is:

[tex]r=k[A]^2[/tex]

The integrated rate law:

[tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]

That is obtained from the integration of:

[tex]\frac{d[A]}{dt}=-k[A]^2[/tex]

And of course, since the slope equals the rate constant, its value is:

[tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]

Regards.

Answer:

[tex]Kp=0.0386[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]

For which the equilibrium expression is:

[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]

Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:

[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]

And the total pressure:

[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]

Yet it is 748 torr, for which the extent is:

[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]

Therefore, Kp turns out:

[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]

Best regards.

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2

Answer:

Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).

Explanation:

Heat of hydrogenation of alkenes is a measure of the stability of carbon-carbon double bonds.

In general, the lower the value of the heat of hydrogenation the more stable the double bond of the alkene.

Also, heat of hydrogenation of alkenes always have a negative value.

Answer:

Percentage mass of Tin = 96.3%

Percentage mass of oxygen = 6.40%

Explanation:

The product of the reaction is an oxide of tin.

Assuming all of the 0.500 g sample of tin reacted with oxygen to produce the oxide:

Mass of oxide = 0.534 g

Mass of tin present in the oxide = 0.500 g

Mass of oxygen in the oxide = 0.534 g of oxide - 0.500 g Sn = 0.034 g O

Percentage composition = mass of element/mass of compound × 100%

Percentage composition of Sn = 0.500 g/0.534 g × 100 = 93.6% Sn

Percentage composition of oxygen = 0.034 g/0.534 g × 100 = 6.40%

Answer:

pH = 4.79

Explanation:

The pH of the acetic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.76

pH = 4.76 + log [sodium Acetate] / [Acetic Acid]

Where [] can be taken as moles of each specie.

Thus, to find pH of the buffer we need to calculate molesof acetic acid and sodium acetate.

Initial moles:

Initial moles of acetic acid and sodium acetate are:

500mL = 0.500L ₓ (0.60moles / L) = 0.30 moles of both acetic acid and sodium acetate

Moles after reaction:

Now, 0.010 moles of NaOH are added to the buffer reacting with acetic acid, CH₃COOH, producing more acetate ion, as follows:

NaOH + CH₃COOH → CH₃COO⁻ + H₂O

That means after reaction moles of both species are:

Acetic acid: 0.30mol - 0.010mol (Moles that react) = 0.29 moles

Acetate: 0.30mol + 0.010mol (Moles produced) = 0.31 moles

Replacing in H-H equation:

pH = 4.76 + log [0.31] / [0.29]

Answer:

The Highly acidic proton joined to one of the carbon in the ALKYNE bond.

(Kindly Check the attachment for the drawing because the solution will need us to draw).

Explanation:

So, let us start by defining some major key terms in this particular Question given above;

(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.

(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.

(3). ALKYNE BOND: it is a C-C joined together by three bonds.

The chemical reaction given in the Question is given in the attachment too.

Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond  is removed irreversibly by NaNH2 base.

The given question is incomplete. The complete question is :

A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.

Answer:  The concentration of barium acetate solution is 0.375 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

[tex]V_s[/tex] = volume of solution in ml

moles of [tex]Ba(CH_3COO)_2[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol[/tex]

Now put all the given values in the formula of molality, we get

[tex]Molarity=\frac{0.125\times 1000}{350ml}[/tex]

[tex]Molarity=0.357M[/tex]

Therefore, the concentration of solution is 0.375 mol/L

Answer:

Final product: cyclopent-1-en-1-ylbenzene

Explanation:

In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).

See figure 1

I hope it helps!

Answer:

d. 2500

Explanation:

In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:

v = k [A]ᵃ [B]ᵇ [C]ⁿ

If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:

v = k [A]ᵃ [B]ᵇ [C]⁰

v = k [A]ᵃ [B]ᵇ×1

Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:

v = k [A]ᵃ [B]ᵇ

v = k [A]ᵃ [B]²

Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:

v = k [A]ᵃ [B]²

v = k [A] [B]²

Relpacing this equation in any experiment (Experiment 5, for example):

20.0 = k [0.200] [0.200]²

2500 = k

That means right answer is:

Answer:

Group 13

Explanation:

You know X has 3 valence electrons, as oxygen has a subscript of 3. This means X has an ionic charge of +3. Group 13 consists mainly of metalloids but it also has metals such as aluminum, which has a +3 charge. If you use aluminum as an example, you know that when combined with oxygen, it forms Al2O3. Group 12 has transition metals that don't have +3 ionic charges, group 14 has metalloids, metals that don't have ionic charges of +3, and nonmetals, and group 2 has metals with ionic charges of +2. Group 13 is the answer.

Answer:

C. rate = k[F₂] [Cl₂O]

Explanation:

Based on the reaction, rate law can be obtained from the initial concentration of reactants thus:

rate = k[F₂]ᵃ [Cl₂O]ᵇ

Where the exponents a and b can be finded doing a experiment changing initial concentrations and seeing how a variation contribute in rate law.

If you analize experiments 1 and 2, the only change is [Cl₂O] (From 0.010 to 0.040, four times more) that changes its concentration in four times. This change produce rate law change from 5x10⁻⁴ to 2.0x10⁻³, also four times. That means the exponent b of [Cl₂O] is 1.

rate = k[F₂]ᵃ [Cl₂O]ᵇ

rate = k[F₂]ᵃ [Cl₂O]¹

Now, comparing experiments 1 and 3, the [F₂] change from 0.05 to 0.10, (Twice), and initial rate change from 5x10⁻⁴ to 1x10⁻³ (Also, twice). That means a = 1 and rate law is:

rate = k[F₂]¹ [Cl₂O]

rate = k[F₂] [Cl₂O]

Thus, right answer is:

Answer:

1. 0.138g of valium would be lethel in the woman

2. 125mg/min is the drip of the patient

Explanation:

1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.

A person that weighs 200lb requires:

200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)

90.72kg × (1.52mg / kg) =

137.9mg ≡

2. The IV contains 1.5g = 1500mg/mL.

If the patient is receiving 5.0mL/h, its rate in mg/h is:

5.0mL/h × (1500mg/mL) = 7500mg/h

Now as 1h = 60min:

7500mg/h × (1h / 60min) =

Answer

mass of aluminum metal= 7 .0497g of Al

Explanation:

current = 21 A

time = 1 hour = 60 X 60 = 3600 s

quantity of electricity passed = current X time = 21X 3600 = 75600 C

Following the electrolysis the below reaction will occur :

Al3+ + 3e- --------> Al

therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al

so 1 C will produce 1/289500 moles of Al

so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al

now 1 mole of aluminium weighs = 27 g/mole

so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g

mass of aluminum metal= 7 .0497 g of Al

The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g

We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:

Current (I) = 21 A

Time(t) = 1 h = 60 × 60 = 3600 s

Q = it

Q = 21 × 3600

Recall:

1 mole of Al = 27 g

1 electron (e) = 96500 C

Thus,

3 electrons = 3 × 96500 = 289500 C

From the balanced equation above,

289500 C of electricity produced 27 g of Al.

Therefore,

75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al

Thus, the mass of the aluminum metal obtained is 7.05 g

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Answer: 9.08 L

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles[/tex]

[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]

According to stoichiometry :

4 moles of [tex]Al[/tex] require  = 3 moles of [tex]O_2[/tex]

Thus 0.54 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 0.54=0.405moles[/tex]  of [tex]O_2[/tex]

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

[tex]V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L[/tex]

Thus 9.08 L of [tex]O_2[/tex] at STP would be required

Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.  

The balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

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:

Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted,   the number of moles of Al that react is calculated as:

[tex]14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles[/tex]

Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?

[tex]amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}[/tex]

amount of moles of O₂= 0.405 moles

On the other side, 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.

Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?

[tex]volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }[/tex]

volume= 9.072 L

Finally, 9.072 L of O₂ at STP would be required.  

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Answer:

[tex]\large \boxed{\text{D. 23.34 min}}[/tex]

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the  half-life

[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]

The half life of the reaction is 23.33 minutes.

We know that for a first order reaction;

ln[A]t = ln[A]o  - kt

A plot of ln[A]t  against time (t) will yield a straight line graph with a slope of -k.

From the question, the slope is -2.97 x 10-2 min-1.

So, -2.97 x 10-2 min-1 = - k

k = 2.97 x 10-2 min-1

The half life of a first order reaction is obtained from;

t1/2 = 0.693/k

t1/2 = 0.693/2.97 x 10-2 min-1

t1/2 = 23.33 minutes

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Answer:

An alkyl halide can undergo SN2 reaction with an amine

Explanation:

The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.

The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).

This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.

The detailed mechanism of this reaction has been attached to this answer.

Answer:

[tex]\boxed{Butyne}[/tex]

Explanation:

Triple Bonds => So it is an alkyne

The suffix used will be "-yne"

4 Carbons => The prefix used will be "But-"

Combining the prefix and suffix, we get:

=> Butyne

Answer:

[tex]\boxed{\mathrm{Butyne}}[/tex]

Explanation:

Alkynes have triple bonds ≡. The molecule has one triple bond.

Suffix ⇒ yne

The molecule has 4 carbon atoms and 6 hydrogen atoms.

Prefix ⇒ But (4 carbons)

The molecule is Butyne.

[tex]\mathrm{C_4H_6}[/tex]

Answer:

KMnO4,H3O^+,75°C

Explanation:

The conversion of cyclohexene to trans-1,2-cylohexanediol is an oxidation reaction. Alkenes are oxidized in the presence of potassium permanganate and acids to yield the corresponding diols.

These diols may also be called glycols. They are molecules that contain two -OH(hydroxyl) groups per molecule. The reaction closely resembles the addition of the two -OH groups of hydrogen peroxide to an alkene.

The bright color of potassium permanganate disappears in this reaction so it can be used as a test for alkenes.

Answer:

See explanation below

Explanation:

Carbon has four electrons in its outermost shell. The CO3^2- anion is found to be in the trigonal planar geometry. For a carbon atom in the trigonal planar geometry, the carbon is sp2 hybridized. This implies that an s orbital mixes with two p orbitals to yield the hybrid orbitals in the ion.

Carbon forms three double bonds to three oxygen atoms using these hybrid sp2 orbitals. Recall that the actual bonding in each C-O linkage lies between that of a pure C-O single bond and C-O double bonds.

Note that there are two p orbitals and one s orbital participating in this hybridization hence three hybrid orbitals are expected to be formed.