Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti or surgical-grade titanium. The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium. What is the empirical formula for surgical-grade titanium

Answer:

Approximately [tex]10.88[/tex].

Explanation:

[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:

[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].

(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)

However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.

At equilibrium:

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].

As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:

[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].

In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].

Again, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:

[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].

Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:

[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].

Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At equilibrium:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].

Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].

That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:

[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 3.35\times 10^{-4}[/tex].

In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:

[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].

(Rounded to two decimal places.)

Answer:

Sn2 mechanism

Explanation:

In this case, our nucleophile is the "OH" on (2R,3R)-3,5-dimethylhexan-2-ol. The alcohol group will attack the [tex]PCl_3[/tex] to produce a new bond between O and P with a positive charge in the oxygen. Additionally, when the OH attacks a Br atom leaves the molecule producing a bromide ion.

In the next step, the bromide ion produced will attack the carbon bonded to the OH that now is bonded to [tex]PCl_2[/tex]. An Sn2 reaction takes place and the substitution would be made in only one step. Due to this, we will have an inversion in the stereochemistry and the absolute configuration on carbon 2 will change from "R" to "S" to produce (2S,3R)-2-bromo-3,5-dimethylhexane.

I hope it helps!

Answer:

Then the hypothesis is proved and becomes a theory.

If not, then another hypothesis should be proposed and tested.

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:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.

Answer:  313.6

Explanation:

To calculate the moles :

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

[tex]\text{Moles of} Fe_2O_3=\frac{450g}{160g/mol}=2.8moles[/tex]

[tex]\text{Moles of} CO=\frac{260g}{28g/mol}=9.3moles[/tex]

[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]

According to stoichiometry :

1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex]  of [tex]CO[/tex]

Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.

As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]

Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]

Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]

Theoretical yield of liquid iron is 313.6 g

Answer:

11445.8years

Explanation:

Half-life of carbon-14 = 5720 years

First we have to calculate the rate constant, we use the formula :

Answer:

B) Ionic

Explanation:

Answer:

Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.  

With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:  

CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.  

Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

Note the following:

Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.

Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:

[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]

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Answer:

c = 250.58 J/kg/[tex]^{0}C[/tex]

Explanation:

The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.

Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.

Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;

c = Q ÷ (mΔθ)

  = 1303 ÷ (1.0 × 5.2)

  = 1303 ÷ 5.2

  = 250.58 J/kg/[tex]^{0}C[/tex]

The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].

Answer:

0.25

Explanation:

Answer:

1. False

2. True

3. True

4. True

5. True

6. True

7. True

8. False

9. True

Explanation:

Density is a physical property since its measurement does not involve any chemical process.

Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.

Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.

The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.

When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.

Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.

The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.

10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.

If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.

Answer:

B!

Explanation:

I got it right in class!

Answer:

F, V, V , V, F

Explanation:

1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".

2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.

3 - ...

4 - ...

5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:

Solido:

 São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.

Liquido:

 São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.

Hibridos:

 O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.

 Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.

Iônica:

 Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.

Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl

Answer:

heptan-2-one

Explanation:

In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.

The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.

See figure 1

I hope it helps!

Answer:

D- poor conductor

Explanation:

metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.

it has a low density and is ductile.

Answer: Poor conductor

Explanation:

Answer:

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

Explanation:

a) The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Enantiometic excess is used to measure the optical purity of an enantiomeric mixture.

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) An enantiomeric excess of 90.2% for (R)-carvone indicates that it's actual percentage is 90.2% more than the percentage of its enantiomeric partner, (S)-carvone, in the mixture.

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

Hope this Helps!!!

a) Specific Rotation of (R)-carvone = -61°

b) The enantiomeric excess of (R)-carvone in the mixture = 90.2%

c) The percentage of (S)-carvone in the mixture = 4.9%

The specific rotation of the enantiomer of a substance is given simply as the negative of the specific rotation of that substance.

Hence, the specific rotation of (R)-carvone is simply the negative of the specific rotation of (S)-carvone.

Specific Rotation of (R)-carvone = -(61°) = -61°

b) Calculation for Enantiomeric excess:

The enantiomeric excess is given mathematically as

ee% = (Observed rotation × 100)/(Specific rotation)

Hence, to calculate the enantiomeric excess of (R)-carvone,

Observed rotation of the mixture = -55°

Specific Rotation of (R)-carvone = -61°

ee% = (-55×100)/(-61) = 90.16% = 90.2%

c) Calculation of percentage:

Let the percentage of (R)-carvone in the mixture be x

Let the percentage of (S)-carvone in the mixture be y

x + y = 100

x - y = 90.2

2x = 190.2

x = (190.2/2) = 95.1%

y = 100 - x = 100 - 95.1 = 4.9%

Hence, the percentage of (R)-carvone in the mixture = 95.1%

The percentage of (S)-carvone in the mixture = 4.9%

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When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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Answer:

6.5 mL

Explanation:

Step 1: Write the balanced reaction

Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O

Step 2: Calculate the reacting moles of nitric acid

25.0 mL of 0.50 M  nitric acid react.

[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]

Step 3: Calculate the reacting moles of calcium hydroxide

The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol

Step 4: Calculate the volume of calcium hydroxide

To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.

[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]

Answer:76 grams

Explanation:

Fe+H₂SO₄-->FeSO₄+H₂

For one mole of Fe we get 1 mole of feso4, therefore for 0.5 moles of Fe we get 0.5 moles of feso4.

The molar mass of feso4 is AFe+AS+4AO(A is atomic mass)

56+32+4*16=152grams/mole

Now, we need to multiply the number of moles by the molar mass to get the mass that reacts

152*0.5=76 grams

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.

Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.

CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

The mass to charge ratio =136

Answer:

I think the answer is " Night vision glasses detect Infrared" energy emitted from cooling objects.

Explanation:

Answer:

-55.9kJ/mol is the change in enthalpy of the reaction

Explanation:

In the reaction:

HCl(aq) + NaOH(aq) → H₂O(l) + NaCl

Some heat is released per mole of reaction.

To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol /  L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol /  L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, HCl is limiting reactant and moles of reaction are moles of limiting reactant, 0.0635 moles

Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)

Replacing:

Q = 4.18J/g°C×100g×8.49°C

Q = 3548.8J of heat are released in the reaction

Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:

ΔH = Heat / mol of reaction

ΔH = -3548.8J / 0.0635 moles of reaction

Negative because is released heat.

ΔH = -55887J / mol

ΔH =

The heat of reaction is  -54.7 kJ/mol.

The equation of the reaction is;

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Number of moles of HCl = 50/1000 L × 1.27 M = 0.064 moles

Number of moles of NaOH = 50/1000 L × 1.32 M = 0.066 moles

The limiting reactant is HCl

Total volume of solution = 100mL

Total mass of solution = 100 g

Temperature rise = 8.49°C

Heat capacity of solution = 4.18 J/g⋅°C

Using;

H = mcdT

m = mass of solution

c = heat capacity of solution

dT = temperature rise

H = 100 g ×  4.18 J/g⋅°C × 8.49°C = 3548.82 J

The heat of reaction = -ΔH/n = -(3.5kJ/0.064 moles)

= -54.7 kJ/mol

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Answer:

The reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C

Explanation:

Hello,

In this case, for the given chemical reaction, we can write the law of mass action (equilibrium expression) as shown below:

[tex]Kc=\frac{[CH_3C_5H_9]}{[C_6H_{12} ]}[/tex]

Thus, since Kc < 1, we can conclude there are more moles of cyclohexane at equilibrium (denominator is greater than numerator), therefore, the reaction will shift leftwards, towards the formation of more cyclohexane at 25 °C.

Best regards.

Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

Answer:

Oxygen and Chlorine

Explanation:

Covalent bonds involve the sharing of electrons between nonmetals.

Answer:

oxygen (O) and chlorine (Cl)

Explanation:

cuz i said so

Answer:

26.5 kD  

Explanation:

Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.56 - ( given ). This is only one part of the information we are given / can conclude in this case ....

i = van’t Hoff factor = 1 for a protein molecule,

R = gas constant = 62.36 L torr / K-mol,

T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K

( Known initially ) ∏ = osmotic pressure = 0.56 torr

..... besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.

_____

Substitute derived / known values to solve for M ( moles / liter ) -

∏ = iMRT

⇒ 0.56 = ( 1 )( M )( 62.36 )( 298 )

⇒ 0.56 = M( 18583.28 )

⇒ M = 0.56 / 18583.28 ≈ 0.00003013461 ....

_____

We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -

M = mol / l

⇒ 0.00003013461 = 0.020 / 25 mL ( 0.025 L ),

0.020 / 0.025 = 0.8 g / L

⇒ 0.8 g = 0.00003013461 moles,

molar mass = 0.8 g / 0.00003013461 moles = 26,548 g / mol = 26.5 kD