In the following reaction: Mg + 2HCl → MgCl2 + H2 How many liters of H2 would be produced if you started with 24.3 g of Mg?

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  

Answer:

Explanation:

In weight/volume (w/v) terms,

   1 ppm = 1g m-3 = 1 mg L-1 = 1 μg mL-1

200 mL = 0.2 L

15 / 0.2 mg L-1 =75 ppm

The concentration in ppm of a solution which is prepared by dissolving in 15mg of NaCl in 200ml water is 75 mg/.,

ppm stand for 'part per million' and it is used to define the concentration of any substance as mass of any substance present in per liter of volume of solution, its unit for measurement is mg/L.

Given that, mass of NaCl = 15mg

Volume of solution = 200mL = 0.2L

Concentration in ppm will be calculated as:
ppm = 15/0.2 = 75mg/L

Hence ppm concentration of NaCl is 75 mg/L.

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Answer:

5.643 kJ

Explanation:

The quantity of heat released or absorbed by a substance (Q) is given by the equation:

Q = mcΔT

Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.

Given that:

m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C

ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C

The quantity of heat is:

Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J

Q = 5.643 kJ

Answer:

diffusion

Explanation:

Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration in response to a concentration gradient. A concentration gradient simply means a difference in concentration.

Diffusion occurs in solids,liquids and gases. Diffusion is fastest in gases and slowest in solids. Diffusion of solid particles may take very many years while diffusion of gases takes a few milliseconds depending on the mass of the gas.

In materials, atoms and molecules also move from one part of the material to another. This is also refereed to as diffusion.

Answer:

(a) appended underneath is the inorganic ion shaped in the reaction and the mechanism of its formation  

(b) 2-butyne framed as a minor product is appeared in the connection. It is shaped when the monosodium subordinate of dimethylsulphoxide gets a hydrogen from the propyne and reacts again with monosodium methylsulphoxide.  

(c) The major product framed when diethylsulphoxide is utilized, would be butyne and minor product would be 3-hexyne.

Explanation:

attached below is diagram

Answer:

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

Explanation:

Double Bond => An Alkene molecule

So, the suffix will be "-ene"

7 Carbons => So, we'll use the prefix "Hept-"

Combining the suffix and prefix, we get:

=> Heptene

Answer:

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

Explanation:

Alkenes have double bonds. The molecule has one double bond.

Suffix ⇒ ene

The molecule has 7 carbon atoms and 14 hydrogen atoms.

Prefix ⇒ Hept (7 carbons)

The molecule is Heptene.

[tex]\mathrm{C_7H_{14}}[/tex]

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is [tex]C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}[/tex]

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.

Correct option = b.

The reaction given in Option A is not a displacement reaction. In Displacement reaction functional group of one reactant is replaced by the functional group of the another reactant.

Displacement reaction:

In this reaction functional group of one reactant is replaced by the functional group of the another reactant.

[tex]\bold { Zn(s) + FeCl_2(aq) \rightarrow ZnCl_2(aq) + Fe(s).}[/tex]

In the above reaction Zinc does not any functional group to exchange with iron chloride.

Therefore, the reaction given in Option A is not a displacement reaction.

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Answer:

24.6g of CO₂ is theoretical yield

Explanation:

The reaction of 8.00g of octane with 38.9g of oxygen.

The reaction of octane with oxygen is:

C₈H₁₈(l) + 25/2O₂ → 9H₂O + 8CO₂

1 mole of octane reacts with 25/2 moles of oxygen to produce 8 moles of CO₂

Theoretical yield is the amount of carbon dioxide formed assuming a yield of 100%. To calculate theoretical yield, first, we need to find limiting reactant and, with the chemical reaction, we can obtain the theoretical moles of CO₂ produced and its mass to obtain theoretical yield.

Limiting reactant:

Moles octane (Molar mass: 114.23g/mol) in 8.00g:

8.00g × (1mol / 114.23g) = 0.0700 moles octane.

Moles oxygen (Molar mass: 32g/mol) in 38.9g:

38.9g × (1mol / 32g) = 1.2156 moles oxygen.

For a complete reaction of 1.2156 moles of O₂ there are necessaries:

1.2156 moles O₂ ₓ (1mol C₈H₁₈ / 25/2 moles O₂) = 0.0973 moles octane

As we have just 0.0700 moles,

Moles and mass of carbon dioxide:

As limiting reactant is octane, 0.0700 moles of C₈H₁₈ will produce:

0.0700mol C₈H₁₈ × (8 moles CO₂ / 1 mol C₈H₁₈) = 0.56 moles of CO₂ are theoretically produced. In mass (Molar mass CO₂ = 44.01g/mol):

0.56moles CO₂ × (44.01g / mol) =

-Theoretical yield because we are assuming all octane is reacting. In real life, never happens like that-

Answer:

Approximately [tex]24\; \rm g[/tex] (at most.)

Explanation:

Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:

[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].

Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:

Calculate the formula mass of the reactants and of the product:

Calculate the quantity (in number of moles of formula units) of each reactant:

Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].

Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].

In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.

On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].

Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].

Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].

Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].

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:

The correct answer is B. Analysis

Explanation:

Error bars are part of the statistical analysis in the scientific method. Once the scientist have collected the data, he or she proceed to the data analysis. A very common way of comparing the data variability is to use error bars in ghaphical representations. From these bars, it can be estimated the error of a determination and experimental groups are compared.

The error bars would be used during B. Analysis.

A blunders bar is a line through a factor on a graph, parallel to one of the axes, which represents the uncertainty or variant of the corresponding coordinate of the point. In IB Biology, the error bars most often represent the same old deviation of an information set.

Blunders bars can be used to examine visual quantities if various other situations preserve. This can decide whether or not differences are statistically sizable. Mistake bars can also propose the goodness of match of a given characteristic, i.e., how well the function describes the facts.

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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:

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:

The answer is 6 because the p sublevel holds 3 orbitals and since each orbital can hold 2 electrons, the answer is 3 * 2 = 6.

Answer:

6 electrons

Explanation:

Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. So the answer is 6 electrons.

Answer:

B) Ionic

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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Answer:

Helium

Explanation:

The speed of the molecules in a gas is directly proportional to the temperature of the gas and inversely proportional to molar mass of the gas.

This implies that when the temperature of a sample of gas is increased, the speed of the gas molecules is increased accordingly.

At a given constant temperature, the molar mass of the gas is inversely proportional to its average molecular speed. This means that the greater the molecular mass of the gas the lesser the average speed of its molecules.

Oxygen has a greater molecular mass than helium hence it will have a lesser average molecular speed compared to helium.

The gas molecule which has the greater average speed is: Helium molecules because they are less massive.

Given the following data:

According to the kinetic-molecular theory, the average speed of gas molecules (particles) is highly dependent on temperature and the molar mass of a gas.

This ultimately implies that, the average speed of gas molecules (particles) is directly proportional to the absolute temperature of an ideal gas and inversely proportional to molar mass of the gas.

At a constant temperature, the higher the molar mass of a gas, the lower would be its average speed and vice-versa.

Hence, helium molecules would have the greater average speed at a constant temperature of 22°C because it is less massive and has a lower molar mass in comparison with oxygen gas.

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Answer:

[tex]M=0.763\frac{mol}{L}=0.763M[/tex]

Explanation:

Hello,

In this case, as the osmotic pressure (π) is widely known as a colligative property, we can see that the solution in this case is formed by water and tree sap, that is mathematically defined by:

[tex]\pi =iMRT[/tex]

Thus, since tree sap is a covalent substance that is nonionizing, we can infer its van't Hoff factor to be 1, therefore, for the given osmotic pressure and temperature, we can compute the molar concentration (in molar units mol/L) as follows:

[tex]M=\frac{\pi }{RT} =\frac{19.1atm}{0.082\frac{atm*L}{mol*K}*(32+273.15)K} \\\\M=0.763\frac{mol}{L}=0.763M[/tex]

Best regards.

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:

65.47∘C

Explanation:

Specific heat capacity, c = 0.128 J/(g⋅∘C)

Initial temperature = 20.0 ∘C

Final temperature = ?

Mass = 52.4 g

Heat = 305 J

All these variables are related by the following equation;

H = m c ΔT

ΔT = H /  mc

ΔT = 305 / (52.4 * 0.128)

ΔT = 45.47∘C

ΔT = Final Temperature - Initial Temperature

Final temperature =  ΔT + Initial temperature

Final temperature = 45.47∘C + 20.0 ∘C = 65.47∘C

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:

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:

The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.

Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.      

HBr will always be the side product of the bromine reactions along with the major compound.                                      

Explanation:

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:

-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:

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:

0.619 M to 3 significant figures.

Explanation:

1 mole of [tex]NiI_{2}[/tex] - 312.5 g

? mole of [tex]NiI_{2}[/tex] - 2.9 g

= 2.9/312.5

= 0.0928 moles.

Concentration = no. of moles/vol in litres = [tex]\frac{0.0928}{0.150L}[/tex]

= 0.619 M

Answer:

Explanation:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

In this reaction we see that 3 moles of O₂ reacts with one mole of C₂H₄ .

Hence rate of disappearance of O₂ is 3 times faster .

- d [O₂] / dt = - 3 d [ C₂ H₄ ] / dt

Putting the given value

.23 = 3 d [ C₂ H₄ ] / dt

d [ C₂ H₄ ] / dt  = .23 / 3

= .077 M / s

Hence the rate of disappearance of C₂ H₄ is .077 moles / s .

Answer:

a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.

Answer:

Oxygen and Chlorine

Explanation:

Covalent bonds involve the sharing of electrons between nonmetals.

Answer:

oxygen (O) and chlorine (Cl)

Explanation:

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