Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 97. g of octane is mixed with 150. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer:

The coefficient for the hydrochloric acid is 6.

Explanation:

Stoichiometric coefficient are the number written in front of the reactants and products in balanced chemical reaction.

When solid aluminum reacts with hydrochloric acid to give aluminum chloride and hydrogen gas.

[tex]2Al(s)+6HCl(s)\rightarrow 2AlCl_3(aq)+3H_2(g)[/tex]

The coefficient for the aluminum is 2.

The coefficient for the hydrochloric acid is 6.

The coefficient for the aluminum chloride acid is 2.

The coefficient for the hydrogen gas  is 3.

Answer:

The pH of a 0.176 M solution of ammonium chloride is 4.9902.

Explanation:

Given:

Kb for ammonia = 1.8×10⁻⁵

Since ammonium ions are the conjugate acid of ammonia.

Thus, Ka for ammonium ions;

Ka = Kw/Kb = 10⁻¹⁴ / 1.8×10⁻⁵ = 5.5556×10⁻¹⁰

Given concentration of Ammonium chloride (C) = 0.176 M

Thus, for weak acids,

[tex]\left[H^+ \right]=\sqrt{K_a\times C}[/tex]

[tex]\left[H^+ \right]=\sqrt{5.5556\times 10^{-10}\times 0.176}[/tex]

[tex]\left[H^+ \right]=0.9777\times 10^{-5}[/tex]

pH is:

[tex]pH\ of\ the\ solution=-log\left[H^+ \right][/tex]

[tex]pH\ of\ the\ solution=-log\left(0.9777\times 10^{-5} \right)[/tex]

pH = 4.9902

Thus,

The pH of a 0.176 M solution of ammonium chloride is 4.9902.

The pH of a 0.176 M solution of ammonium chloride at 25 degrees Celsius is 11.01, calculated by using the equilibrium equation of the hydrolysis of the ammonium ion and taking into account the neutrality of the reaction.

The question requires calculating the pH of a 0.176 M solution of ammonium chloride (NH4Cl), which is formed from the neutralization of the weak base ammonia (NH3) by the strong acid hydrochloric acid (HCl).

To find the pH, it is necessary to find the concentration of the hydronium ion [H3O+]} in the solution. The neutrality of the reaction means that when NH3 reacts with HCl, it forms its conjugate acid NH4+, which hydrolyzed in water. The equilibrium equation for the hydrolysis NH4+ + H20 <-> NH3 + H3O+ and knowing the value of Kb for ammonia (1.8×10−5), you can calculate Ka for the ammonium ion, which helps calculate [H3O+]. Using the expression Kw = Ka x Kb, since Kw is 1.0 x 10^-14 at 25 degrees Celcius, you can find Ka = Kw / Kb, which equals 5.56 x 10^-10. Using the hydrolysis reaction equilibrium equation, Ka = [NH3][H3O+] / [NH4+], and knowing [NH4+] = 0.176M (from the problem statement), you isolate [H3O+] and find it equals 9.79 x 10^-12. Finally, using the pH=-log10[H3O+], calculate the pH to be 11.01.

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Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

Volume of water = [tex]4.50\times 10^2ml=0.45L[/tex]

The decomposition of KClO  will be :

[tex]KClO\rightarrow K^++ClO^-[/tex]

Now the further reaction with water [tex](H_2O)[/tex] to give,

[tex]ClO^-+H_2O\rightarrow HClO+OH^-[/tex]

First we have to calculate the pOH.

[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8[/tex]

Now we have to calculate the [tex]OH^-[/tex] concentration.

[tex]pOH=-\log [OH^-][/tex]

[tex]3.8=-\log [OH^-][/tex]

[tex][OH^-]=1.58\times 10^{-4}M[/tex]

Now we have to calculate the base dissociation constant.

Formula used : [tex]K_b=\frac{K_w}{K_a}[/tex]

Now put all the given values in this formula, we get :

[tex]K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}[/tex]

Now we have to calculate the concentration of [tex]ClO^-[/tex].

The equilibrium constant expression of the reaction  is:

[tex]K_b=\frac{[OH^-][HClO]}{[ClO^-]}[/tex]

As we know that, [tex][OH^-]=[HClO]=1.58\times 10^{-4}M[/tex]

[tex]2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}[/tex]

[tex][ClO^-]=0.0999M[/tex]

Now we have to calculate the moles of [tex]ClO^-[/tex].

[tex]\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole[/tex]

As we know that, the number of moles of [tex]ClO^-[/tex] are equal to the number of moles of KClO.

So, the number of moles of KClO = 0.0449 mole

Now we have to calculate the mass of KClO.

[tex]\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO[/tex]

[tex]\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g[/tex]

Therefore, the mass of potassium hypochlorite is, 4.1 grams.

To find the amount of potassium hypochlorite required, we first calculate the concentration of [OH-] ions from the given pH, then use this to calculate the amount of hypochlorite ions required. Our calculations yield an approximate amount of 0.15g which, among the provided options, the closest is 0.032g.

In this problem, we are asked to determine the mass of potassium hypochlorite that must be added to water to give a solution with a pH of 10.20. Potassium hypochlorite is a weak base, and the formula for pH is pH = 14 - pOH. Since pOH is the negative log of the concentration of OH- ions, we can rearrange to find [OH-].

First, find the pOH: pOH = 14 - pH = 14 - 10.2 = 3.8. Then find the [OH-]: [OH-] = 10^-pOH = 10^-3.8. This gives the concentration of hypochlorite ions (OCl-) in solution because in water it dissociates as KOCl -> K+ + OCl-.

Using the molar mass of potassium hypochlorite, we can find the mass that must be added. The molar mass is given as 90.6 g/mol. So to find the mass, we multiply the volume of the water (which must be in liters, so 4.5 * 10^-2 L) by the [OH-] and then by the molar mass of potassium hypochlorite. Thus, mass = volume * [OH-] * molar mass =~ 0.15g. Hence the closest answer is 0.032g.

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Answer:

The formula of the hydrate of the sodium sulfate is :[tex]Na_2SO_4.7H_2O[/tex]

Explanation:

Mass of hydrated sodium sulfate = 40.15 gram

Mass of completely dehydrated sodium sulfate = 21.27 gram

Mass of water molecules present in hydrated sodium sulfate =x

40.15 gram =  21.27 gram + x (Law of conservation of mass)

x  = 40.15 gram - 21.27 gram = 18.88 g

Moles of water = [tex] \frac{18.88 g}{18 g/mol}= 1.04mol[/tex]

Moles of sodium sulfate =[tex]\frac{21.27 g}{142.04 g/mol}=0.1497 mol[/tex]

Whole number ratio of sodium sulfate and water:

Sodium sulfate =[tex]\frac{0.1497 mol}{0.1497 mol}=1[/tex]

Water =[tex]\frac{1.04 mol}{0.1497 mol}=6.9 a\\rox 7[/tex]

The formula of hydrate be [tex]Na_2SO_4.7H_2O[/tex]

So, the formula of the hydrate of the sodium sulfate is :[tex]Na_2SO_4.7H_2O[/tex]

The formula of the hydrate, the mass of water lost is calculated by subtracting the mass of anhydrous Na₂SO₄ from the original hydrate mass. The number of moles of each substance is then calculated and the ratio of moles of water to moles of  Na₂SO₄ is approximately 7, giving a hydrate formula of Na₂SO₄·7H₂O.

The formula of the hydrate of Na₂SO₄, we first determine the mass of the water lost during heating by subtracting the mass of the anhydrous Na₂SO₄ from the original mass of the hydrate:

Mass of water lost = original mass - mass of anhydrous Na₂SO₄

Mass of water lost = 40.15 grams - 21.27 grams = 18.88 grams

Now, to find the number of moles of Na₂SO₄ and H₂O, we use their molar masses (Na₂SO₄ = 142.04 g/mol, H₂O = 18.01 g/mol):

Moles of Na₂SO₄ = 21.27 grams / 142.04 g/mol = 0.1498 moles

Moles of H₂O = 18.88 grams / 18.01 g/mol = 1.048 moles

The mole ratio of H₂O to Na₂SO₄ is found by dividing the moles of H₂O by the moles of Na₂SO₄:

Mole ratio = moles of H₂O / moles of Na₂SO₄

Mole ratio = 1.048 moles / 0.1498 moles ≈ 7

Therefore, the empirical hydrated compound formula is  Na₂SO₄·7H₂O.

Answer: [tex]3.9\times 10^{23}[/tex] iron atoms

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

[tex]1 molecule of [tex]Fe_2O_3[/tex] contains= 2 atoms of iron

[tex]1 mole of [tex]Fe_2O_3[/tex] contains=[tex]2\times 6.023\times 10^{23}=12.05\times 10^{23}[/tex]  atoms of iron

thus 0.32 moles of [tex]Fe_2O_3[/tex] contains=[tex]\frac{12.05\times 10^{23}}{1}\times 0.32=3.9\times 10^{23}[/tex]  atoms  of iron

Thus the sample would have [tex]3.9\times 10^{23}[/tex] iron atoms.

Answer:

True

Explanation:

Limiting reactant - the reactant which get completely consumed in a chemical reaction , is known as the limiting reactant .

As, the concentration of limiting reactant after the completion of the reaction will be zero , hence, it is used to determine the concentration of other reactants .

For example,

for a general reaction -

A + B ---> 3C

Assuming B to be the limiting reactant ,

hence, the concentration of C and A can be determined as -

1 mol of B can give 3 mol of C and 1 mol of A is used for the reaction.

Answer: Amount of tetraphosphorus decoxide formed is 38.60g, amount of potassium chloride formed is 33.99g and amount of red phosphorus consumed in the reaction is 81.48 g.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ....(1)

Given mass of potassium chlorate = 56.0 g

Molar mass of potassium chlorate = 122.55 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of potassium chlorate}=\frac{56.0g}{122.55g/mol}=0.456mol[/tex]

For the given chemical reaction:

[tex]10KClO_3+12P\rightarrow 3P_4O_{10}+10KCl[/tex]

Red phosphorus is given in excess . So, it is considered as an excess reagent and potassium chlorate is considered as a limiting reagent.

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 3 moles of tetraphosphorus decoxide

So, 0.456 moles of potassium chlorate will react with = [tex]\frac{3}{10}\times 0.456=0.136moles[/tex] of tetraphosphorus decoxide

Calculating the mass of tetraphosphorus decoxide by using equation 1, we get:

Molar mass of tetraphosphorus decoxide = 283.886 g/mol

Moles of tetraphosphorus decoxide = 0.136 moles

Putting values in equation 1, we get:

[tex]0.136mol=\frac{\text{Mass of tetraphosphorus decoxide}}{283.886g/mol}\\\\\text{Mass of tetraphosphorus decoxide}=38.60g[/tex]

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 10 moles of potassium chloride

So, 0.456 moles of potassium chlorate will react with = [tex]\frac{10}{10}\times 0.456=0.456moles[/tex] of potassium chloride

Calculating the mass of potassium chloride by using equation 1, we get:

Molar mass of potassium chloride = 74.55 g/mol

Moles of potassium chloride = 0.456 moles

Putting values in equation 1, we get:

[tex]0.456mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=33.99g[/tex]

By Stoichiometry of the reaction:

10 moles of potassium chlorate reacts with 12 moles of red phosphorus.

So, 0.456 moles of potassium chlorate will react with = [tex]\frac{12}{10}\times 0.456=2.631moles[/tex] of red phosphorus

Calculating the mass of red phosphorus by using equation 1, we get:

Molar mass of red phosphorus = 30.97 g/mol

Moles of red phosphorus = 2.631 moles

Putting values in equation 1, we get:

[tex]2.631mol=\frac{\text{Mass of red phosphorus}}{30.97g/mol}\\\\\text{Mass of red phosphorus}=81.48g[/tex]

Hence, amount of tetraphosphorus decoxide formed is 38.60g, amount of potassium chloride formed is 33.99g and amount of red phosphorus consumed in the reaction is 81.48 g.

Answer:

The final temperature of the system is 42.46°C.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))[/tex]

where,

c = specific heat of water= [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of water sample with 100 °C= 50.0 g

[tex]m_2[/tex] = mass of water sample with 13.7 °C= 100.0 g

[tex]T_f[/tex] = final temperature of system

[tex]T_1[/tex] = initial temperature of 50 g of water sample= [tex]100^oC[/tex]

[tex]T_2[/tex] = initial temperature of 100 g of water =[tex]13.7^oC[/tex]

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

[tex]50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))[/tex]

[tex]T_f=42.46^oC[/tex]

The final temperature of the system is 42.46°C.

Answer:

The correct answer is C where K=[CO2]^2[H2O]^3/[C2H5OH][O2]^3

Explanation:

Equilibrium constant (K) is expressed as the product of all the products divided by the product of all reactants. the coefficient in front of each reactants or products are used as their exponents. Solids and liquids are considered to be 1 hence they are not shown, but in this case everything is in gaseous form thus they are all expressed in the equilibrium formula formula.

Final answer:

The equilibrium constant for the combustion of ethanol is expressed using the balanced chemical equation, providing the correct formula for K from the given options.

Explanation:

Equilibrium constant (K) for the combustion of ethanol:

Write the balanced equation: C₂H₅OH(g) + 3O₂(g) ⇌ 2CO₂(g) + 3H₂O(g).

Express the equilibrium constant: K = [CO₂]²[H₂O]³ / [C₂H₅OH][O₂]³.

Hence, the correct choice is Option C.

Answer:

a)905,89 atm of pressure will be generated at 298K.

b)The sap of tree can rise upto 9,357.84 meters.

Explanation:

a)

Effective concentration of sap = c = 37 M

Osmotic pressure generate at 298K = [tex]\pi [/tex]

Temperature ,T = 298 K

[tex]\pi =cRT[/tex]

[tex]\pi =37 mol/L\times 0.08216 atm L/mol L\times 298 K[/tex]

[tex]\pi =905.89 atm[/tex]

905,89 atm of pressure will be generated at 298K across the endodermis root membrane.

b)

Given that 1.0 at of pressure raises the volume of water upto height of 10.33 m

Then 905.89 atm of pressure will raise the height of water upto:

[tex]\frac{10.33}{1.0}\times 905.89 m=9,357.84 m[/tex]

The sap of the tree can rise upto 9,357.84 meter.

The osmotic pressure generated across the root membrane of a tree with a 37 Molar sap concentration at 298K is 903.62 atm. This pressure could theoretically raise the sap to a height of about 9334.39 meters, though such a height is unrealistic in actual tree physiology due to factors like structural limits.

To calculate the osmotic pressure generated at 298K across the endodermis root membrane for a tree sap with an effective concentration of 37 Molar, we use the formula
c = iMRT, where:

i is the vant Hoff factor (which is 1 for sugars)

M is the molarity of the solution (37 M)

R is the ideal gas constant (0.08216 L.atm/mol.K)

T is the temperature in Kelvins (298K)

Plugging in the numbers, we get:
c = (1)(37 M)(0.08216 L.atm/mol.K)(298K) = 903.62 atm

To determine how high the sap can rise given an osmotic pressure of 1.0 atm can raise a column of water 10.33 meters, we set up a proportion:
(1 atm / 10.33 m) = (903.62 atm / x m)

Solving for x gives us x = 903.62 atm * 10.33 m/atm ≈ 9334.39 meters. Therefore, theoretically, the sap could rise to a height of approximately 9334.39 meters due to the generated osmotic pressure. However, this scenario is not realistic due to limitations such as hydrostatic pressure and structural integrity of the tree.

Osmotic pressure and root pressure are essential in understanding the transport of water and nutrients in plants, particularly in how they work together with transpiration and cohesive forces in water molecules to move sap through the xylem vessels from the roots to the leaves.

Answer and Explanation:

Calorie is the unit of heat energy . There are 2 units with the same name 'calorie' which is widely used.

'The amount of heat energy required to increase the temperature of 1 gram of water by mass by [tex]1^{\circ}C[/tex] or 1 K is known as small calorie or gram calorie'.

Another one is large calorie which can be defined as :

'The amount of heat energy required to make arise in temperature of water 1 kg by mass by [tex]1^{\circ}C[/tex] or 1 K is known as large calorie or  kilcalorie and is represented as Cal or kcal'.

After the adoption of SI system, thee units of the metric system cal, C or kilocal are considered deprecated or obsolete with the SI unit for heat energy as 'joule or J'

1 cal = 4.184 J

1C or 1 kilocal = 4184 J

Calorimeter constant:

Calorimeter constant, represented as '[tex]C_{cal}[/tex]' is used to quantify the heat capacity or the amount of heat of a calorimeter.

It can be calculated by ther given formula:

[tex]{\displaystyle C_{cal}}={\frac {\Delta {H}}{\Delta {T}}}}}[/tex]

where,

[tex]{\Delta {T}}[/tex] = corresponding temperature change

[tex] {\Delta {H}[/tex] = enthalpy change

Its unit is J/K or J/1^{\circ}C[/tex] which can be convertyed to cal/1^{\circ}C[/tex] by dividing the calorimeter constant by 4.184 or 4184 accordingly.

A calorie is the amount of heat energy required to raise the temperature of one gram of water by one degree Celsius, equating to 4.184 Joules. The calorimeter constant is the heat capacity of the calorimeter, calculated by dividing the amount of heat absorbed by the temperature increase during a calorimetry experiment.

A calorie is a conventional unit of heat, defined as the quantity of heat energy required to raise the temperature of one gram of water by one degree Celsius. This definition of a calorie aligns specifically to the temperature change from 14.5°C to 15.5°C.To convert calories into Joules, the SI unit of heat, a constant of 4.184 Joules equals one calorie.

The calorimeter constant (also known as the heat capacity of the calorimeter) is vital in calorimetry, the process of measuring heat energy changes in a chemical system. It is specified in units of cal/°C. To calculate the calorimeter constant for your calorimeter, you would have to know the mass of the calorimeter and the temperature change observed during an experiment. The constant is calculated by dividing the amount of heat absorbed (in calories, calculated using known quantities and specific heat values of substances involved) by the temperature increase.

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Answer: The equations are given below.

Explanation:

For the given options:

The molecular equation for this follows:

[tex]AgNO_3(aq.)+NaI(aq.)\rightarrow AgI(s)+NaNO_3(aq.)[/tex]

Ionic form of the above equation follows:

[tex]Ag^+(aq.)+NO_3^-(aq.)+Na^+(aq.)+I^-(aq.)\rightarrow AgI(s)+Na^+(aq.)+NO_3^-(aq.)[/tex]

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]Ag^+(aq.)+I^-(aq.)\rightarrow AgI(s)[/tex]

The molecular equation for this follows:

[tex]Ba(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+BaSO_4(s)[/tex]

Ionic form of the above equation follows:

[tex]Ba^{2+}(aq.)+2NO_3^-(aq.)+K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2K^+(aq.)+2NO_3^-(aq.)+BaSO_4(s)[/tex]

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)[/tex]

The molecular equation for this follows:

[tex]2NH_4NO_3(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+H_2SO_4(aq.)+NH_3(g)[/tex]

Ionic form of the above equation follows:

[tex]2NH_4^+(aq.)+2NO_3^-(aq.)+2K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)+SO_4^{2-}(aq.)+2K^+(aq.)+2NO_3^-(aq.)[/tex]

As, nitrate, potassium and sulfate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]2NH_4^+(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)[/tex]

The molecular equation for this follows:

[tex]3LiCl(aq.)+Al(NO_3)_3(aq.)\rightarrow 3LiNO_3(aq.)+AlCl_3(s)[/tex]

Ionic form of the above equation follows:

[tex]3Li^+(aq.)+3Cl^-(aq.)+Al^{3+}(aq.)+NO_3^-(aq.)\rightarrow 3Li^+(aq.)+3NO_3^-(aq.)+AlCl_3(s)[/tex]

As, lithium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]Al^{3+}(aq.)+3Cl^-(aq.)\rightarrow AlCl_3(s)[/tex]

Hence, the molecular and ionic equations are given above.

Hey there!

500 mg of protein is present in 100 mL of solvent as per the concentration 0.5 mg/mL or 500 g/mL ,

So, 250 mg (0.25 g) of serving food need to be added to 100 mL solvent in order to prepare 50 mg of protein/100 mL solution.

Dilution factor = initial amount of protein / final amount of protein

= 6 g / 0.05 g = 120

Hope this helps!

When 5.60 mol of ethane is burned, 11.20 mol of CO2 are produced.

The balanced equation for the combustion of ethane (C2H6) is 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g). From this equation, we can see that for every 2 moles of ethane burned, 4 moles of CO2 are produced. Therefore, to calculate how many moles of CO2 are produced when 5.60 mol of ethane is burned, we can use the ratio:

2 moles of ethane : 4 moles of CO2 = 5.60 mol of ethane : x moles of CO2

Solving for x, we find that x = (5.60 mol of ethane) x (4 moles of CO2) / (2 moles of ethane) = 11.20 mol of CO2.

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5.60 moles of ethane produce 11.20 moles of carbon dioxide when burned in excess oxygen, based on the 2:4 molar ratio from the balanced combustion equation. This demonstrates the stoichiometric calculation for the reaction. Thus, burning ethane produces twice as many moles of CO₂.

To solve this, we need to use the balanced chemical equation for the combustion of ethane:

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

This equation tells us that 2 moles of ethane produce 4 moles of carbon dioxide. Therefore, the ratio of moles of ethane to moles of carbon dioxide is 2:4, or 1:2.

If 5.60 moles of ethane (C₂H₆) are burned, we can use this ratio to find the moles of CO₂ produced:

Thus, burning 5.60 moles of ethane produces 11.20 moles of carbon dioxide.

The correct answer is: b. raise the freezing point.

When pressure increases, the freezing point of a substance generally increases. This phenomenon is known as the colligative property of freezing point elevation. In a system where solid and liquid phases coexist, an increase in pressure tends to favor the denser phase.

Since the liquid phase is denser than the solid phase, an increase in pressure would promote the transition from solid to liquid, causing the freezing point to rise. This effect is commonly observed in many substances and is utilized in various applications, such as in the preservation of food through high-pressure processing.

Therefore, under increased pressure, the freezing point of the substance would be raised, making option b the correct choice.

Answer: The equilibrium concentration of [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] are 0.0266 M.

Explanation:

The chemical equation for the ionization of calcium chromate follows:

[tex]CaCrO_4\rightleftharpoons Ca^{2+}+CrO_4^{2-}[/tex]

The expression for equilibrium constant is given as:

[tex]K_c=\frac{[Ca^{2+}][CrO_4^{2-}]}{[CaCrO_4]}[/tex]

We are given:

[tex]K_c=7.1\times 10^{-4}[/tex]

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Let the equilibrium concentration for [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] be 'x'

Putting values in above equation, we get:

[tex]7.1\times 10^{-4}=x^2\\\\x=0.0266M[/tex]

Hence, the equilibrium concentration of [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] are 0.0266 M.

hey there!:

C2H4

1 c=c ->611

4 C-H -> 4*414=1656

=> Ha=2267 kj

H2 :  H-H 436

Hb = 436

C2H6

1 C-C 347

6 C-H 6*414=2484

=> Hc=2831

H=(Ha+Hb)-Hc=2267+436-2831 = -128kj

Answer A

Answer:

[tex]\Delta H_{rxn}^o=-128kJ[/tex]

Explanation:

Hello,

In this case, the standard enthalpy of reaction could be computed via the bond energies when both broken or made as shown below:

[tex]\Delta H_{rxn}^o=\Delta H_{broken}+\Delta H_{made}[/tex]

In this manner, we infer that at the reactants for ethene, [tex]C_2H_4[/tex] a double bond between carbons is broken as well as a bond between hydrogens (such values turn out positive). Furthermore, a single bond between carbons and two single bonds between carbon and hydrogen are made (such values turn out negative), in such a way, we develop the aforesaid equation to obtain:

[tex]\Delta H_{rxn}^o=(611kJ+436kJ)+(-347kJ-2*414kJ)\\\Delta H_{rxn}^o=-128kJ[/tex]

Best regards.

Answer : The solubility at [tex]25^oC[/tex] is, 0.0658 M

Explanation :

According to the Henry's Law, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

[tex]S\propto P[/tex]

or,

[tex]\frac{S_1}{S_2}=\frac{P_1}{P_2}[/tex]

where,

[tex]S_1[/tex] = initial solubility of carbon dioxide gas = 0.0347 M

[tex]S_2[/tex] = final solubility of carbon dioxide gas = ?

[tex]P_1[/tex] = initial pressure of carbon dioxide gas = 775 torr

[tex]P_2[/tex] = final pressure of carbon dioxide gas = 1470 torr

Now put all the given values in the above formula, we get the final solubility of the carbon dioxide gas.

[tex]\frac{0.0347M}{S_2}=\frac{775\text{ torr}}{1470\text{ torr}}[/tex]

[tex]S_2=0.0658M[/tex]

Therefore, the solubility at [tex]25^oC[/tex] is, 0.0658 M

Using Henry's Law, the solubility of carbon dioxide in water at 25 °C would double from 0.0347 M at 775 Torr to 0.0694 M at 1470 Torr.

To determine the solubility of carbon dioxide at 25 °C and 1470 Torr, we can use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. Given that the solubility of carbon dioxide is 0.0347 M at 775 Torr, we can calculate its solubility at 1470 Torr (which is double the initial pressure) by simply doubling the initial solubility. Therefore, at 25 °C and 1470 Torr, the solubility of carbon dioxide in water would be 0.0694 M.

Answer:The statements b,c,d would lead to increased formation of more hydrogen gas and statements a, e and f would  lead to unchanged hydrogen concentration

Explanation:

Lechateliers principle can be used here to determine the effect  of changes observed in the system.

Lechateliers principle states that if  any reaction at equilibrium  is subjected to change in concentration, temperature and pressure or even in reaction conditions  then the equilibrium of the reaction would shift in such a way so that it can oppose the change .

So if any disturbance is caused to a reaction  at equilibrium hence  the equilibrium of reaction would shift in such a way so that it can counter balance the change caused to the reaction.

The above reaction is following:

C(s)+H₂O(g)→CO(g)+H₂(g)

The enthalpy change  of this reaction is positive and hence the reaction is endothermic in nature.

So the given changes would lead to the following  results:

a The addition of more amount of carbon C(s) would not lead to any further formation of hydrogen because carbon is added in solid state and Hydrogen gas is in gaseous state so the equilibrium for this given reaction would only change on addition of gaseous reactants as that would only lead to change in concentration.

b Since H₂O(g) is in gaseous state and a reactant and hence the addition of  H₂O(g) that is more reactant would lead to more formation of hydrogen gas according to lechatelier principle. The equilibrium would shift in such a way so that it can decrease the concentration of added H₂O(g) hence it would form H₂(g).

c Since the above reaction is endothermic in nature hence increasing the temperature of reaction would also shift the equilibrium of reaction towards more formation of H₂(g) that is in forward direction.

d When we increase the volume of reaction mixture that is we are increasing the amount of reactants hence the reaction would shift towards more formation of hydorgen gas.

e The catalyst does not change the position of equilibrium and hence no shift in position of equilibrium would be observed.So amount of hydrogen gas formed would remain unchanged.

f The addition of inert gas would not lead to any change to the reaction and equilibrium would be unaffected. Hence the formation of hydrogen gas would remain unchanged.

The study of chemicals and the bond is called chemistry. When the amount of the reactant and the product get equal is said to be equilibrium.

The correct answer is b, c, d would lead to increased formation of more hydrogen gas, and statements a, e and f would lead to unchanged hydrogen concentration.

According to the principle can be used here to determine the effect of changes observed in the system.

So if any disturbance is caused to a reaction at equilibrium hence the equilibrium of reaction would shift in such a way so that it can counterbalance the change caused to the reaction.

The reaction is as follows:-

[tex]C(s)+H_2O(g)---->CO(g)+H_2(g)[/tex]

The enthalpy change of this reaction is positive and hence the reaction is endothermic in nature.

So the given changes would lead to the following results:

A. The addition of more amount of carbon C(s) would not lead to any further formation of hydrogen because carbon is added in solid-state. Hydrogen gas is in the gaseous state so the equilibrium for this given reaction would only change on the addition of gaseous reactants as that would only lead to a change in concentration.

B. Since H₂O(g) is in a gaseous state and a reactant and hence the addition of  H₂O(g) that is more reactant would lead to more formation of hydrogen gas according to the principle. The equilibrium would shift in such a way so that it can decrease the concentration of added H₂O(g) hence it would form H₂(g).

C . Since the above reaction is endothermic in nature hence increasing the temperature of the reaction would also shift the equilibrium of reaction towards more formation of H₂(g) that is in the forward direction.

D. When we increase the volume of the reaction mixture that is we are increasing the number of reactants hence the reaction would shift towards more formation of hydrogen gas.

E. The catalyst does not change the position of equilibrium and hence no shift in the position of equilibrium would be observed. So the amount of hydrogen gas formed would remain unchanged.

F. The addition of inert gas would not lead to any change to the reaction and equilibrium would be unaffected. Hence the formation of hydrogen gas would remain unchanged.

Hence, the correct answer is mentioned above.

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Answer:The chemical shift (δ) is 6.88ppm.

Explanation:

We have the following data :

Absorption frequency of the proton in bromoform=2065Hz

Frequency of the NMR spectrometer(instrument)=300MHz

The formula for calculating the chemical shift (δ) in PPM is:

chemical shift (δ)={[Frequency of proton(Hz) -Frequency of reference(Hz]÷Frequency of NMR spectrometer(MHz }

Using the formula for chemical shift we can calculate the value of chemical shift (δ) in ppm

chemical shift (δ) ={[2065Hz-0]÷300×10⁻⁶}

chemical shift (δ) =6.88×10⁶

chemical shift (δ)=6.88ppm for a 300MHz NMR spectrometer

The chemical shift (δ) is a ratio of frequency absorbed by proton with that of NMR spectrometer frequency hence the chemical shift value would remain same what ever NMR spectrometer frequency we use. Chemical shift basically tells us about the position of signal with respect to the reference compound of TMS(δ=0).

chemical shift (δ) is measured in  

So the value of (δ) is same for any spectrometer used.

The chemical shift (δ) for a 500MHz NMR spectrometer used would also be 6.88PPM.

Alternatively since the frequency of proton absorbed is directly related to the magnetic field applied that is the  frequency of NMR spectrometer hence:

Let the frequency of proton absorbed in 300MHZ=V₁=2065

Let the frequency of proton absorbed in 500MHZ=V₂=?

frequency of proton absorbed∝Applied magnetic field(Frequency of NMR spectrometer)

So V₁/V₂=[300×10⁶]/[500×10⁶]

V₂=[2065×500]÷300

V₂=3441HZ

For a 500 MHz proton the frequency of absorption would be 3441MHz

using this frequency we can calculate chemical shift (δ) using above formula:

(δ)=[3441-0]/[500×10⁻⁶]

(δ)=6.88PPM

Hence we obtain the same value of chemical shift in both the spectrometers.

This answer explains how to calculate the chemical shift of a nucleus using NMR spectroscopy when switching between different instrument frequencies.

Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful tool used by chemists to analyze the structure of molecules. In NMR spectroscopy, the chemical shift is a measure of the resonance frequency of a nucleus compared to a standard.

To find the chemical shift on a 500-MHz instrument, we use the formula: Chemical shift on New Instrument = (Resonance Frequency on Old Instrument) × (New Instrument Frequency) / (Old Instrument Frequency).

Plugging in the values, we get: Chemical shift = 2065 Hz × 500 MHz / 300 MHz = 3441.67 Hz.

Therefore, the chemical shift of the CHBr₃ proton on a 500-MHz NMR instrument would be 3441.67 Hz.

Based on the data provided, it appears that the metal M with a molar mass of 27.0 g/mol reacts with hydrochloric acid to produce hydrogen gas, indicating the metal is likely aluminum. The reaction equation would be 2Al + 6HCl → 2AlCl3 + 3H2. The oxide of aluminum is Al2O3 and the sulfate is Al2(SO4)3.

To find the equation for the reaction of the metal M with hydrochloric acid, we should first understand the typical reaction that occurs, which is a metal reacting with an acid to produce a salt and hydrogen gas:

M + 2HCl → MCl2 + H2

Given the data, we can calculate the moles of hydrogen gas (H2):

n(H2) = PV / RT

Where P is the pressure in atmospheres (741 mmHg is equivalent to 741/760 atm), V is the volume (0.303 L), R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (17°C = 290 K).

Once we have the moles of H2, we can find the ratio of metal M to H2 in the reaction and then use the molar mass of M to determine how many moles of M reacted, which should be half the moles of H2 produced since the metal is likely to be in Group 1 or 2. Lastly, using the molar mass of the metal (27.0 g/mol), we can confirm the stoichiometry.

For the formulas of the oxide and sulfate of M:

In summary, the molar mass given and typical valence of metals may suggest that metal M is likely aluminum (Al), hence the possible chemical reactions would be:

2Al + 6HCl → 2AlCl3 + 3H2

Formulas for aluminum compounds:

Oxide of Al: Al2O3

Sulfate of Al: Al2(SO4)3

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The metal M is aluminum (Al), and the Balanced reaction is:  [tex]2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2[/tex] the Oxide is: [tex]\text{Al}_2\text{O}_3[/tex], and Sulfate is [tex]\text{Al}_2(\text{SO}_4)_3[/tex].

To determine the reaction and write the formulas for the oxide and sulfate of the metal  M , we first need to identify the metal and its valency based on the data provided.

1. Determine the number of moles of hydrogen gas liberated:

Using the ideal gas law,  PV = nRT :

-  P  is the pressure in atm. To convert from mmHg to atm:  [tex]P = \frac{741 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.974 \, \text{atm}[/tex].

-  V  is the volume in liters:  V = 0.303 L.

-  R  is the ideal gas constant:  R = 0.0821 L·atm/(mol·K.

-  T  is the temperature in Kelvin:  [tex]T = 17^\circ \text{C} + 273.15 = 290.15 \, \text{K}[/tex].

Now, solve for  n , the number of moles of  H₂ :

[tex]n = \frac{PV}{RT} \\\\ n = \frac{(0.974 \, \text{atm})(0.303 \, \text{L})}{(0.0821 \, \text{L\ . atm/(mol\ . K)})(290.15 \, \text{K})} \\\\ n = \frac{0.295 \, \text{L\ . atm}}{23.81 \, \text{L\ . atm/mol}} \\\\ n \approx 0.0124 \, \text{mol}[/tex]

2. Determine the number of moles of metal  M  reacted:

Given the mass of the metal and its molar mass:

[tex]\text{Moles of metal} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.225 \, \text{g}}{27.0 \, \text{g/mol}} = 0.00833 \, \text{mol}[/tex]

3. Determine the valency of the metal  M :

From the reaction with hydrochloric acid, each mole of metal  M  produces hydrogen gas. The general reaction is:

[tex]M + xHCl \rightarrow MCl_x + \frac{x}{2}H_2[/tex]

Since 0.00833 moles of  M  produced 0.0124 moles of  H₂:

[tex]0.00833 \, \text{mol} \times \frac{x}{2} = 0.0124 \, \text{mol} \\\\ x = \frac{0.0124 \times 2}{0.00833} \\\\ x \approx 2.98 \approx 3[/tex]

So, the valency of metal  M  is 3.

4. Write the chemical reaction:

The balanced equation for the reaction is:

[tex]2M + 6HCl \rightarrow 2MCl_3 + 3H_2[/tex]

5. Write the formulas for the oxide and sulfate of  M :

- The formula for the oxide of  M :

[tex]M_2O_3[/tex]

- The formula for the sulfate of  M :

[tex]M_2(SO_4)_3[/tex]

Given that the metal has a valency of 3, and the molar mass of 27 g/mol corresponds to aluminum (Al), the formulas for the oxide and sulfate of  M  (Aluminum) are:

- Oxide:  [tex]\text{Al}_2\text{O}_3[/tex]

- Sulfate: [tex]\text{Al}_2(\text{SO}_4)_3[/tex]

Thus, the metal  M  is aluminum (Al), and the balanced reaction, oxide, and sulfate formulas are:

- Balanced reaction:

[tex]2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2[/tex]

- Oxide:

[tex]\text{Al}_2\text{O}_3[/tex]

- Sulfate:

[tex]\text{Al}_2(\text{SO}_4)_3[/tex]

Answer: Strong electrolytes:[tex](HNO_{3})[/tex] Nitric acid, [tex](Ca(OH)_{2})[/tex])Calcium Hydroxide and (KCl) potassium chloride

Weak electrolytes: [tex](CH_{3}COOH)[/tex]Acetic acid and [tex](CH_{3}NH_{2})[/tex] Methyl amine

Non-electrolytes:[tex](C_{2}H_{5}OH)[/tex]Ethanol and [tex](C_{6}H_{12}O_{6})[/tex]Glucose

Explanation: Electrolytes are those compounds which can conduct electricity when dissolved in any polar solvent.

Strong electrolytes are those compounds which completely ionise when dissolved in polar solvent and hence produce ions in solution . So greater the capacity of an compound to ionize itself greater number of ions would be present in solution and hence greater will be the capacity of the solution to conduct electricity.

Ionic compounds like [tex](HNO_{3})[/tex] Nitric acid ,(KCl) Potassium chloride and [tex](Ca(OH)_{2})[/tex])Calcium hydroxide are completely ionized  when dissolved in polar solvent so these compounds are strong electrolytes.

Weak electrolytes are those  compounds which undergo partial ionization when dissolve in polar solvents . So they are not able to produce more ions in the solution and hence the conductivity of a solution containing weak electrolytes is low.

[tex](CH_{3}COOH)[/tex]Acetic acid and [tex]CH_{3}NH_{2}[/tex]Methyl amine are partially ionized when dissolved in polar solvent so these electrolytes are weak electrolytes.

Non-electrolytes are those compounds which are not at all ionized in the polar solvent and they remain as molecules itself even if they are dissolved.

[tex](C_{2}H_{5}OH)[/tex]Ethanol and  [tex](C_{6}H_{12}O_{6})[/tex]Glucose do not ionize when dissolved in polar solvent and remain as molecules itself so the solutions of these compounds will not have ions and hence they would be unable to conduct electricity.

so

Specific rotation of a liquids can be determined from concentration, length of column and its optical rotation. The specific rotation of the solution in 5 cm length column and 3.5 g in 1 mL is -12.57°.

Optical rotation of a solution is the angle of rotation of a plane polarized light which passes through the solution.  If the solution is in 1 decimeter column with 1g/l of concentration then it is called specific rotation.

The plane polarized light have orientation to either one direction left or right. The plane polarized light to left is leavo and that of right is called dextro.

The equation for specific rotation using density in g/ml and length of column in decimeter is as shown below:

specific rotation  = [tex]\frac{\alpha }{d \times l}[/tex]

Where, l is the column length and d is density, alpha is the optical rotation.

Apply thr given values to the above equation.

5 cm = 0.5 decimeter.

specific rotation = - 0.022 /(0.5 × 3.5/1 ml)

                           = -12.57°.

Hence, the specific rotation of the solution in 5 cm length column and 3.5 g in 1 mL with an optical rotation of -0.022 is -12.57°.

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Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

[tex]\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ][/tex]

Where,

[tex]\Delta H_v[/tex] is the Heat of vaoprization (J/mol)

[tex]T_b[/tex] is the normal boiling point of the gas (K)

[tex]T_c[/tex] is the Critical temperature of the gas (K)

[tex]P_c[/tex] is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

[tex]T_b=307.4\ K[/tex]

[tex]T_c=466.7\ K[/tex]

[tex]P_c=36.4\ bar[/tex]

Applying the above equation to find heat of vaporization as:

[tex]\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ][/tex]

[tex]\Delta H_v=26400 J/mol[/tex]

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

[tex]\Delta H_v=26.4 kJ/mol[/tex]

Option C is correct

Answer:

The answer is B. the hydrophilic end has the hydrocarbon chain.

Explanation:

That answer is wrong because soap and detergent form a bilayer with hydrophilic and hydrophobic ends.

the hydrophilic end interacts with polar compound such as water because that end is ionic and charged, it is composed of carbon and Oxygen and forms dipole interaction and electrostatic interactions.

the hydrophobic end is the non polar end, it is composed of carbon and hydrogen chain (Hydrocarbon) which are non-polar due to equal electronegativity between carbon and hydrogen.

that is why answer B is not true, the hydrophilic end does not have the hydrocarbon chain, the hydrocarbon chain is on the hydrophobic end.

Answer: The value of [tex]\Delta S^o[/tex] for the given reaction is -620.1 J/Kmol.

Explanation:

Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.

Mathematically,

[tex]\Delta S_{rxn}=\sum [n\times S^o_{products}]-\sum [n\times S^o_{reactants}][/tex]

For the given chemical equation:

[tex]2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)[/tex]

Taking the standard entropy of formation for the following:

[tex]S^o_{C_2H_6}=229.6Jmol^{-1}K^{-1}\\S^o_{CO_2}=213.6Jmol^{-1}K^{-1}\\S^o_{H_2O}=69.9Jmol^{-1}K^{-1}\\S^o_{O_2}=205Jmol^{-1}K^{-1}[/tex]

Putting values in above equation, we get:

[tex]\Delta S^o_{rxn}=[(4\times S^o_{CO_2})+(6\times S^o_{H_2O})]-[(2\times S^o_{C_2H_6})+(7\times S^o_{O_2})][/tex]

[tex]\Delta S^o=[(4\times 213.6)+(6\times 69.9)]-[(2\times 229.6)+(7\times 205)]=-620.1Jmol^{-1}K^{-1}[/tex]

Hence, the [tex]\Delta S^o[/tex] of the reaction is [tex]-620.1Jmol^{-1}K^{-1}[/tex]

The standard entropy change for the complete combustion of ethane is -620.8 J/K.mol.

The standard entropy change is equal to the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants.

Let's consider the combustion of ethane.

2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l)

The standard entropy change (ΔS°) is:

ΔS° = n(products) × S°products - n(reactants) × S°reactants

where,

ΔS° = 4 × 213.74 J/mol.K + 6 × 69.91 J/mol.K - 2 × 229.60 J/mol.K - 7 × 205.14 J/mol.K = -620.8 J/K.mol

The standard entropy change for the complete combustion of ethane is -620.8 J/K.mol.

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Answer:

Fertilizers are chemicals that kill plants.

Explanation:

Fertilizers are not pesticides. Fertilizers helps to improve plant yield by supplying nutrients needed for plant growth and food formation. When fertilizers are applied to plants, the nutrient pool increases and plants can readily grow without any deficiency resulting from lack of one mineral or the other.

Herbicides are used to kill plants.

Answer:

It would not affect the volume of the base needed to reach the equivalence point

Explanation:

Once the unknown acid has been prepared and placed in the conical flask ready for titration, adding water to it cannot alter how much base is needed to neutralise it. Adding water to the acid merely dilutes the acid i.e if 20 parts of acid were in say 100 parts of water now 20 parts of acid are in 300 parts of water because you added more water. You will still need the exact amount of base to neutralise the 20 particles that are still in the conical flask awaiting titration.

Adding water does not make acid particles increase or decrease, it just means the base particles will have to collide with a lot more water particles to interact with the acid particles which will slow down the reaction. So the rate of the reaction can be affected, not volume of base needed to reach equivalence point.