calculate the solubility of oxygen in water at 25C when the total external pressure is 1 at and the mole fractionof oxygen in the air is 0.2

Answer:

                      %age Yield  =  66.05 %

Explanation:

                     In order to check the efficiency of a reaction the %age is calculated so that it can be concluded that either the conditions provided for certain reactions are favourable or not because at large scales reactions with greater %age yields are favoured as it results in high product quantity and hence, greater economical advantages.

                   The percent yield is given as;

                  %age Yield  =  Actual Yield / Theoretical Yield × 100

Putting values,

                   %age Yield  =  253 g / 383 g × 100

                   %age Yield  =  66.05 %

Hence, for the given reaction the success of the reaction is only 65% hence, steps can be made to improve this yield by modifying the reaction conditions.

Answer:

16.8 g

Explanation:

We are told than burning one mol of ethanol releases 1368 kJ. Now we are trying to find how much ethanol has to be burned, in grams, to release 500 kJ

We use ratios

-1368 kJ : 1 mole

-500 kJ :      x

Then you cross multiply

-1368x = -500

       x = 0.3655 mol

mass = number of moles * molar mass

        = 0.3655 mol * 46.07 g/mol

        = 16.8 g

The mass of ethanol ([tex]C_2H_5OH_{(l)}[/tex]) that must be burned to supply 500 kJ of heat is 16.84 grams.

Given the following data:

We know that the molar mass of ethanol ([tex]C_2H_5OH_{(l)}[/tex]) is equal to 46.07 g/mol.

To calculate the mass of ethanol ([tex]C_2H_5OH_{(l)}[/tex]) that must be burned to supply 500 kJ of heat:

By stoichiometry:

1 mole of ethanol = 1368 kJ of heat

X mole of ethanol = 500 kJ

Cross-multiplying, we have:

[tex]1368 \times X = 500\\\\X = \frac{500}{1368}[/tex]

X = 0.3655 moles

Now, we can determine the mass of ethanol required:

[tex]Mass = molar \;mass \times number\;of\;moles\\\\Mass = 46.07 \times 0.3655[/tex]

Mass = 16.84 grams

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If matter is uniform throughout, cannot be separated into other substances by physical processes, but can be decomposed into other substances by chemical processes, it is compound. Thus option D is correct.

Anything that takes space and mass called as Matter. It may be atom or any other element made up of space and mass only.

Atoms bond together form molecule, compound and matter such as solid, liquid and gas.

we cannot break atom as it is the smallest unit of a matter. Atom is made up of Electrons, protons, and neutrons and size is around 100 picometers.

A compound is a complex of molecule which are made up of number of atoms by forming a bond called chemical bonds.

Depending on the the bond pattern of atoms, compounds have different bonds such as covalent bond, ionic bond, metallic bonds, coordinate covalent bonds.

Thus option D is correct.

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Answer:

C.) perpendicular

Explanation:

A particle with an electric charge experiences the maximum deflecting force when it is positioned perpendicular to the magnetic field.

Answer:

Explanation:

This question we will solve by calculations based on the stoichiometry of the balanced chemical equation which gives us all the information required to know the quantities produced and reacted based on their molar ratios.

First we will need the molecular weights of the reactants to calculate the number of moles of each reactant and determine if there is a limiting reagent, and from there we can learn about the moles and masses of the products.

2 C3H6(g) + 2 NH3(g) + 3 O2(g) ⇒ 2 C3H3N(g) + 6 H2O(g)

MW C3H6 : 42.08 g/mo                    MW C3H3N : 53.06 g/mol

MW  NH3 : 17.03 g/mol                      MW  H2O : 18.02 g/mol

MW  O2: 32 g/mol

Moles of reactants:

Convert the masses given to grams since we have the molar masses in grams. The number of moles, n, is calculated by dividing the mass into the molecular weight.

n C3H6 = ( 1.16 Kg x 1000 g/ Kg ) / 42.08 g/mol = 27.56 mol

n NH3 = ( 1.65 kg x 1000 g/ /Kg ) / 17.03 g/mol   = 96.89 mol

n O2 =  ( 1.78 Kg x 1000 g/ Kg ) /  32 g/mol = 55.63 mol

from the stoichiometry of the reaction we know propylene and ammonia react  2: 2  so  propylene is the limiting reagent:

( 2 mol NH3 / 2 mol C3H6 )x  27.56 mol C3H6 = 27.56 mol NH3 (required to react with the 27.56 mol C3H6 and we have plenty ( 96.98 mol )

The stoichiometry of the reaction also confirms that O2 is in excess:

(  3 mol O2 / 2 mol C3H6 ) x 27.56 mol C3H6 = 41.34 mol O2  (required to react completely with 27.56 mol C3H6 ).

(a) Again from the balanced chemical reaction we know the mol proportions reactants to product, thus mol C3H3N ( 1: 1 ) produced:

( 2 mol C3H3N / 2 mol C3H6 ) x 27.56 mol C3H6 = 27.56 mol C3H3N

The mass of acrylonitrile will be given by multiplying the molecular weight of the mol produced assuming a 100 % yield:

55.12 g/mol x  27.56 mol  = 1,519 g = 1.51  Kg

(b) The calculation to obtain the mass of water will be performed in a similar manner:

( 6 mol H2O / 2 mol C3H6 ) x 27.56 mol C3H6 = 82.68 mol H2O produced

82.68 mol x 18 g/mol = 1,488 grams = 1.49 Kg

(c) The mass of O2 left will be obtained from the number of moles in excess:

mol O2 originally present = 1.78 x 1000 g/Kg / 32 g/mol = 55.62 mol

mol O2 in excess = mol O2 initially - mol reacted

from above we know 41.34 mol are required to react with our limiting reagent, C3H6 :

mol O2 in excess = 55.62 mol - 41.34 mol = 14.29 mol

mass oxygen in excess = 32 g/mol x 14.29 mol = 457.12 g = 0.457 Kg

Answer:

Remain the same

Explanation:

The law of mass conservation states that matter cannot be created nor destroyed but can be converted from one form to another.

Essentially, what Lavoisier was trying to proof is that by heating the mixture, after all the change the mass still remains. That was why he used a sealed flask. If the flask was not sealed, it probably would have been that some of the mass will escape as vapor to the atmosphere which might be difficult to account for

Answer:

[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]

[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]

[tex]pH = 0.119[/tex]

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

[tex]n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)[/tex]

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

[tex]n_{H^{+} } from HCl = (5.00)(0.093)[/tex]

[tex]n_{H^{+} } from HCl = 0.465 mol[/tex]

[tex]n_{H^{+} } from HNO_{3} = (8.00)(0.037)[/tex]

[tex]n_{H^{+} } from HNO_{3} = 0.296 mol[/tex]

[tex]n_{H^{+}(total) } = 0.296 + 0.465[/tex]

[tex]n_{H^{+}(total) } = 0.761 mol[/tex]

For molar concentration of hydrogen ions:

[tex][H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}[/tex]

[tex][H^{+}] = \frac{0.761}{1.00}[/tex]

[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

[tex]K_{w} = [H^{+} ][OH^{-} ][/tex]

[tex][OH^{-}]=\frac{Kw}{[H^{+}] }[/tex]

[tex][OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }[/tex]

[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]

The pH of the solution can be measured by the following formula:

[tex]pH = -log[H^{+} ][/tex]

[tex]pH = -log(0.761)[/tex]

[tex]pH = 0.119[/tex]

The type of interparticle force that has the greatest influence on the physical properties of each substance are identified. Factors that result in a stronger ionic bond overall are listed.

The type of interparticle force that has the greatest influence on the physical properties of each substance are as follows:

The factors that result in a stronger ionic bond overall are:

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Cl2 and CH4 experience London dispersion forces, NH2OH has hydrogen bonding, PCl3 has dipole-dipole interactions, and CaCl2 and KI demonstrate ionic bonding. Greater absolute charges and smaller ion sizes boost the strength of ionic bonds.

The substances listed – Cl2, NH2OH, PCl3, CH4, CaCl2, and KI – have different types of interparticle forces that affect their physical properties. Cl2 and CH4 are nonpolar with London dispersion forces; NH2OH has hydrogen bonding; PCl3 has dipole-dipole interactions. CaCl2 and KI are ionic compounds, where the primary interparticle force is ionic bonding.

The factors that enhance the strength of ionic bonds are usually the greater absolute charges on the ions and the smaller size of the ions. Larger charges mean that the electrostatic attraction between the ions is stronger, leading to a stronger ionic bond. Smaller ions, on the other hand, can get closer together, which also intensifies the electrostatic attraction and hence increases the strength of the ionic bond.

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Answer:

30.10%

Explanation:

The mass of water is:

50.00 g − 34.95 g = 15.05 g

So the percentage of water is:

15.05 g / 50.00 g × 100% = 30.10%

Final answer:

The experimental percentage of water in copper(II) sulfate pentahydrate, after heating a sample and measuring the mass loss, is calculated to be 30.10%.

Explanation:

Experimental Percentage of Water in a Hydrate

The question involves calculating the percentage of water in a hydrate, specifically copper(II) sulfate pentahydrate, which is a classic chemistry experiment. To determine the percentage of water, you have to compare the mass of water lost upon heating to the original mass of the hydrate.

First, calculate the loss of mass due to heating: 50.00 grams (initial mass) - 34.95 grams (final mass) = 15.05 grams of water lost.

Percentage of water in the hydrate is found by the formula:
Mass% of H₂O = (Mass of water lost / Original mass of hydrate) * 100%.
Substituting the known values gives us: Mass% of H₂O = (15.05 g / 50.00 g) * 100% = 30.10%.

The experimental percentage of water in the copper(II) sulfate pentahydrate is therefore 30.10%, which is determined by this experimental process of heating and weighing.

Answer:

A.Brønsted-Lowry base

Explanation:

Answer:

B:Arrhenius base

Explanation:

Acccording  to Svante Arrhenius a base is a substance the dissociates in water to form hydroxide ion.In other words a base is that substance which when dissolved in water,increases the concentration of hydroxide ion.The examples of Arrhenius bases are given below;

NaOH ,KOH,LiOH etc

Dissociation of NaOH in aqueous solution is given as;

NaOH=>Na+ + OH-

Answer:

5118.30485 moles

Explanation:

There are approximately 44.0095 moles of CO2 in 1 gram. So just multiply 44.0095 by  116.3.

Answer:

omework Help. Steno's laws of stratigraphy describe the patterns in which rock layers are deposited. The four laws are the law of superposition, law of original horizontality, law of cross-cutting relationships, and law of lateral continuity.

Explanation:

Answer:

a

Explanation:

Answer:

Vb = 18 L option c)

Explanation:

First, we need to write the titration reaction between the base and the acic, which is the following:

Ba(OH)₂ + H₃PO₄ <-------> Ba₃(PO₄)₂ + H₂O

However this equation is not balanced, we need to balance the equation adding some coefficients to the agents so:

3Ba(OH)₂ + 2H₃PO₄ <-------> Ba₃(PO₄)₂ + 6H₂O

Now that the equation is balanced, as we know this is an acid base titration, we need to calculate the mole ratio between the base and acid so:

moles B / moles A = 3/2

2 moles B = 3 moles A (1)

This is taken from the balanced reaction.

Now, finally we use the relation in titration which is:

moles A = moles B

or simply MaVa = MbVb

If we replace this in the ratio of this reaction we have:

2MbVb = 3MaVa (2)

And from there, we solve for Vb which is the volume of the base:

2 * 2 * Vb = 3 * 4 * 6

4Vb = 72

Vb = 72/4

Vb = 18 L

This is the volume of the base required to titrate this acid

Answer:

150 mL of the 30% alcohol mixture and 30 mL of the pure water will we need to obtain the desired solution.

Explanation:

Let the volume of 30% alcohol used to make the mixture = x L

For 25% alcohol:

C₁ = 25% , V₁ = 180 mL

For 30% alcohol :

C₂ = 30% , V₂ = x L

Using  

C₁V₁ = C₂V₂

25×180 = 30×x

So,  

x = 150 mL

Pure water = 180 mL - 150 mL = 30 mL

150 mL of the 30% alcohol mixture and 30 mL of the pure water will we need to obtain the desired solution.

Answer: Option A. All of these statements are reasonable compromises if you MUST wear these items to lab, but the best and safest practice is to leave them at home and dress for lab intentionally.

Explanation: in the lab., good safety practice is best rather than assumed safety.

Answer:

The answer is A

Answer:

c. ) polyethylene with high density plasticizers added to increase density.

Explanation:

High - density polyethylene -

Polyethylene is a crystalline structure , which have a very wide area of application , it is a type of thermoplastic polymer .

It is produced in huge tons every year due to its wide range of application.

Commercially it is produced by the very famous , the Ziegler - Natta catalysts .

In the polyethylene , a high - density polyethylene is added in order increase its density , and hence this type of polyethylene is referred to as high - density polyethylene .

High-density polyethylene (HDPE) is composed of primarily linear, unbranched chains of polyethylene in a close packing arrangement. This structure gives HDPE a higher tensile strength and makes it suitable for stronger and more rigid products.

High-density polyethylene (HDPE) is composed of primarily linear, unbranched chains of polyethylene in a close packing arrangement. Therefore, the correct answer is b). HDPE has a relatively low degree of branching, which allows the polymer chains to pack closely together, resulting in stronger intermolecular forces and a higher tensile strength compared to more highly branched polymers like low-density polyethylene (LDPE). HDPE's characteristics make it suitable for use in products that require strength and rigidity, such as detergent bottles, milk jugs, and water pipes.

In contrast, LDPE has a higher degree of branching, which prevents the polymer chains from packing as tightly, leading to a more flexible material with a lower tensile strength. This type of polyethylene is used for products that need to be more flexible, such as beach balls and plastic bags. HDPE and LDPE are both types of polyethylene, but they have different properties due to their molecular structure and how the polymer chains pack together.

Answer:

Option B.

Explanation:

As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:

2C₂H₆  +  7O₂   →   4CO₂  +  6H₂O

So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.

Then 2 moles of ethane will produce 4 moles of CO₂

The final answer is:

[tex]\[ \Delta S^\circ_{\text{rxn}} = 1716.72 \, \text{J/mol-K} \][/tex]

Let's calculate the energy change of the reaction using the given data:

Reaction equation:

[tex]\[ \text{Sn}(s) + 2\text{Cl}_2(g) \rightarrow \text{SnCl}_4(l) \][/tex]

Given:

[tex]\[ \Delta H^\circ_{\text{rxn}} = -511.3 \, \text{kJ/mol} \]\\Temperature (\( T \)) = \( 25^\circ \text{C} \) = \( 298.15 \, \text{K} \)\\Pressure (\( P \)) = \( 1.00 \, \text{bar} \)\\\\[/tex]

First, we convert[tex]\( \Delta H^\circ_{\text{rxn}} \)[/tex] from kJ/mol to J/mol:

[tex]\[ \Delta H^\circ_{\text{rxn}} = -511.3 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -511300 \, \text{J/mol} \][/tex]

Next, we use the formula for Gibbs Free Energy [tex](\( \Delta G^\circ_{\text{rxn}} \)):[/tex]

[tex]\[ \Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T \Delta S^\circ_{\text{rxn}} \][/tex]

At standard conditions [tex](\( 25^\circ \text{C} \) and \( 1.00 \, \text{bar} \)), \( \Delta G^\circ_{\text{rxn}} \)[/tex] is equal to zero:

[tex]\[ 0 = -511300 \, \text{J/mol} - (298.15 \, \text{K}) \times \Delta S^\circ_{\text{rxn}} \][/tex]

Solving for[tex]\( \Delta S^\circ_{\text{rxn}} \):[/tex]

[tex]\[ \Delta S^\circ_{\text{rxn}} = \frac{-(-511300 \, \text{J/mol})}{298.15 \, \text{K}} \]\[ \Delta S^\circ_{\text{rxn}} = 1716.72 \, \text{J/mol-K} \][/tex]

So, the final answer is:

[tex]\[ \Delta S^\circ_{\text{rxn}} = 1716.72 \, \text{J/mol-K} \][/tex]

1. Convert [tex]\( \Delta H^\circ_{\text{rxn}} \)[/tex]  from kJ/mol to J/mol.

2. Use the formula [tex]\( \Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T \Delta S^\circ_{\text{rxn}} \) and solve for \( \Delta S^\circ_{\text{rxn}} \) at standard conditions where \( \Delta G^\circ_{\text{rxn}} = 0 \).[/tex]

Complete Question:

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, calculate the energy change of the reaction at 25 ∘ C and 1.00 bar . Sn ( s ) + 2 Cl 2 ( g ) ⟶ SnCl 4 ( l ) Δ H ∘ rxn = − 511.3 kJ/mol

The standard internal energy change for the reaction is approximately [tex]\( -511.3 \text{ kJ/mol} \).[/tex]

The standard internal energy change for the given reaction at 25°C and 1.00 bar can be calculated using the standard enthalpy change, H°rxn, provided that the reaction occurs at constant pressure and the only work done is pressure-volume work. Under these conditions, the change in internal energy (U°rxn) can be approximated by the change in enthalpy (Hrxn), because the difference between H°rxn and U°rxn is the product of the pressure, volume change, and the number of moles of gas, which is often small for reactions that do not involve a significant amount of gas.

 The reaction is as follows:

[tex]\[ \text{Sn (s)} + 2\text{Cl}_2 \text{(g)} \rightarrow \text{SnCl}_4 \text{(l)} \][/tex]

 Given:

[tex]\[ \Delta H^\circ_{\text{rxn}} = -511.3 \text{ kJ/mol} \][/tex]

At constant pressure (1.00 bar) and temperature (25°C), the relationship between Hrxn and U°rxn is given by:

[tex]\[ \Delta H^\circ_{\text{rxn}} = \Delta U^\circ_{\text{rxn}} + P\Delta V \][/tex]

 For reactions involving gases, the term [tex]\( P\Delta V \)[/tex] represents the work done by the system on the surroundings due to volume change against the constant external pressure \( P \). Since the number of moles of gas decreases during the reaction (2 moles of Cl2 gas are consumed to form 1 mole of SnCl4 liquid), \( \Delta V \) is negative, and thus \( P\Delta V \) is also negative.

However, for condensed phases (solids and liquids), the volume change is typically small, and the [tex]\( P\Delta V \)[/tex] term is often negligible compared to the enthalpy change. Therefore, for the reaction given, which involves a solid reactant and a liquid product, we can assume that [tex]\( \Delta H^\circ_{\text{rxn}} \approx \Delta U^\circ_{\text{rxn}} \).[/tex]

 Thus, the standard internal energy change for the reaction is approximately equal to the standard enthalpy change:

[tex]\[ \Delta U^\circ_{\text{rxn}} \approx \Delta H^\circ_{\text{rxn}} \][/tex]

[tex]\[ \Delta U^\circ_{\text{rxn}} \approx -511.3 \text{ kJ/mol} \][/tex]

 Therefore, the standard internal energy change for the reaction is approximately [tex]\( -511.3 \text{ kJ/mol} \).[/tex]

 In conclusion, the energy change of the reaction at 25°C and 1.00 bar is:

[tex]\[ \boxed{\Delta U^\circ_{\text{rxn}} \approx -511.3 \text{ kJ/mol}} \][/tex]

This value is a good approximation for the standard internal energy change of the reaction under the given conditions.

 The answer is: [tex]\Delta U^\circ_{\text{rxn}} \approx -511.3 \text{ kJ/mol}.[/tex]

Answer:

a. Heat Energy

b. Temperature

Explanation: Heat energy are transferred from one point to another due to difference in temperature

Answer:

yes because it is a great planet

Explanation:

because of the reason for silicon

Answer:

Silicon-28

Explanation:

To estimate the isotope with the highest percentage of abundance, we look at which of the isotopes have the closest resemblance to the value of the final atomic mass.

From what is given, it can be observed that the isotope with the highest value is that of the 28 because it can be seen that it is large enough to have affected the value of the isotope near it.

Answer:

The answer to your question is  "It will be a high melting point"

Explanation:

Process

1.- Identify the kind of compounds that is Potassium chloride.

Ionic compounds are composed of a metal and a nonmetal.

Covalent compounds are composed of 2 nonmetals.

Potassium chloride is composed of a metal and a nonmetal so, it is  an ionic compound.

2.- Conclude, Potassium chloride has a high melting point because is an ionic compound".

Answer:

High melting point.

Explanation:

Potassium chloride is an ionic compound  so the melting point will be  high.

Answer:

The products are carbon dioxide and water

Explanation:

Step 1: Data given

Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve  O2  as one reactant.

Step 2: The complete combustion of C3H7OH:

For the combustion of 1-propanol, we need O2.

The products of this combustion are CO2 and H2O.

C3H7OH + O2→ CO2 + H2O

On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3

C3H7OH + O2→ 3CO2 + H2O

On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.

C3H7OH + O2→ 3CO2 + 4H2O

On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).

To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.

2C3H7OH + 9O2→ 6CO2 + 8H2O

For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O

The products are carbon dioxide and water

Answer:

Carbon dioxide and water

Explanation:

Answer : The mass of reactant [tex]H_2[/tex] remain would be, 0.20 grams.

Solution : Given,

Moles of [tex]H_2[/tex] = 0.40 mol

Moles of [tex]O_2[/tex] = 0.15 mol

Molar mass of [tex]H_2[/tex] = 2 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]O_2[/tex] react with 2 mole of [tex]H_2[/tex]

So, 0.15 moles of [tex]O_2[/tex] react with [tex]0.15\times 2=0.30[/tex] moles of [tex]H_2[/tex]

From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.

The moles of reactant [tex]H_2[/tex] remain = 0.40 - 0.30 = 0.10 mole

Now we have to calculate the mass of reactant [tex]H_2[/tex] remain.

[tex]\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2[/tex]

[tex]\text{ Mass of }H_2=(0.10moles)\times (2g/mole)=0.20g[/tex]

Therefore, the mass of reactant [tex]H_2[/tex] remain would be, 0.20 grams.

Answer:

The remaining mass of [tex]\rm H_2[/tex] in the reaction is 0.20 grams.

Explanation:

The balanced equation for the reaction will be:

[tex]\rm 2\; H_2 \; + O_2 \rightarrow\; 2\; H_2O[/tex]

1 mole of [tex]\rm O_2[/tex] reacts with 2 moles of [tex]\rm H_2[/tex] to gives 2 moles of [tex]\rm H_2O[/tex].

0.15 moles of [tex]\rm O_2[/tex] reacts with 2 * 0.15 = 0.30 moles of [tex]\rm H_2[/tex]

We have 0.40 moles of [tex]\rm H_2[/tex]

So remaining [tex]\rm H_2[/tex]= 0.40 moles - 0.30 moles

                         = 0.10 moles

Mass = moles * molar mass

Molar mass of [tex]\rm H_2[/tex] = 2 g/mol

Mass of [tex]\rm H_2[/tex] remained in the reaction = 0.10 moles * 2 g/mole

                                                            = 0.20 grams.

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Answer:

Piruvate

Explanation:

Glucose + ATP32 --------> Glucose-6-P32 ----->Fructose-6-P32

Fructose-6-P32 + ATP32 ---------> Fructose-1,6 bi-P32

Fructose-1,6 bi-P32 ----------> Glyceraldehyde-3-P32 + Dihydroxiacetone Phosphate32 (that are in equilibrium)

Glyceraldehyde-3-P32 +NADH +Pi (non radioactive) -----> Glyceraldehyde-1,3-P32 (1,3 biphosphoglycerate where only in C3, there is a P32)

After that, the phosphorus in Carbon1 is donated to ADP to for ATP (non radioactive), and we have 3-Phosphoglycerate (radioactive because its P32), then it's converted to 2-Phosphoglycerate (radioactive), then Phosphoenolpiruvate (radioactive), that donates its P32 to ADP to produce ATP, remaining Piruvate as end product at the end of glucolysis

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O     ⇄   H₃O⁺ +   F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the  [ H₃O⁺ ]  ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1  relation , we will also have [F⁻ ]. The  [ HF ] is given in the question so we have all the information that is needed to  compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻²  M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x  ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴  

(Rounded to two significant figures, the powers of 10 have infinite precision )

To calculate the Ka for hydrofluoric acid, use the given pH to find the hydronium ion concentration, apply the dissociation reaction, and solve for the acid dissociation constant assuming x (the degree of dissociation) is small.

To calculate the acid dissociation constant (Ka) for hydrofluoric acid we can use the formula for pH, which is pH = -log[H+], where [H+] represents the concentration of hydronium ions. Given that the pH is 1.68,

we can find the concentration of H+:

[H+] = 10-pH = 10-1.68

The acid dissociation reaction for HF is:

HF(aq) ⇌ H+(aq) + F-(aq)

Assuming the degree of dissociation is x,

we would have that [H+] = [F-] = x,

and the initial concentration of HF after dissociation would be 0.65 - x.

At equilibrium, we have:

Ka = [H+][F-]/[HF] = x2/(0.65 - x)

Because x is small compared to the initial concentration of HF (0.65 M),

we can approximate this as:

Ka ≈ x2/0.65

Now substituting the calculated [H+] for x and solving for Ka, we get the acid dissociation constant for hydrofluoric acid.

Answer:

The rate constant for a second order reaction is

k = 0.51 dm-3 s-1.

Explanation:

In a regular second-order reaction the rate equation is given by v = k[A][B], if the reactant B concentration is constant then v = k[A][B] = k'[A], where k' the pseudo–first-order rate constant = k[B].

Also 1/|A| = Kt + 1/|Ao|

But Ao = 0.657 M and

A = 0.0981 M also

t = 17.0 s

Therefore

1/| 0.0981 M| =K × 17.0 s + 1/| 0.657 M|

→ 10.19/M = 17K + 1.52/M

10.19/M - 1.52/M = 8.67/M = 17K

K = 8.67M/17s = 0.51 dm-3 s-1.

k = 0.51 dm-3 s-1.

Answer:

1. By Pressure factor: if we double the pressure volume become half of its original

2. 2.14 L

3. 2.15 L

Explanation:

part 1

Data Given:

volume of container change

temperature of remain constant

The pressure doubles

Solution:

This problem can be explained by Boyle's Law that at constant temperature pressure and volume has an inverse relation with each other.

So the volume change due to change in Pressure.

            P1V1 = P2V2

if we consider conditions at STP, as follows

initial volume V1 = 22.42 L

and

initial pressure P1 = 1 atm

if the pressure doubles then

final pressure P2 = 2 atm

Put values in Boyle's law equation

     (1 atm) (22.42L) = (2 atm) (V2)

Rearrange the above equation to find V2

           V2 =    (1 atm) (22.42L) / 2 atm

            V2 = 11.12 L

So it is clear from calculation if we double the pressure volume become half of its original. So its the pressure due to volume become effected and decrease by its increase.

_____________

Part 2

Data Given:

Initial temperature T1= 250 K

final Temperature T2= 350 K

initial volume V1 =  ?

final volume V2 = 3.0 L

Solution:

This problem will be solved by Charles' Law equation at constant pressure

      V1 / T1 = V2 / T2 . . . . . . . . (1)

put values in above equation

      V1 / 250 K = 3.0 L / 350 K

Rearrange the above equation to calculate V1

       V1  = (3.0 L / 350 K) x 250 K

       V1  = (0.0086 L . K) x 250 K

       V1  = 2.14 L

So the initial volume = 2.14 L

_________________

part 3

Data Given:

Initial temperature T1= 20 ºC

Convert Temperature from ºC to Kelvin

T = ºC + 273

T = 20 + 273 = 293 K

final Temperature T2= -5.00 ºC

Convert Temperature from ºC to Kelvin

T = ºC + 273

T = - 5.00 + 273 = 268 K

initial volume V1 =  2.35 L

final volume V2 = ?

Solution:

This problem will be solved by Charles' Law equation at constant pressure

      V1 / T1 = V2 / T2 . . . . . . . . (1)

put values in above equation

      2.35 L / 293 K = V2 / 268 K

Rearrange the above equation to calculate V1

       V2  = (2.35 L / 293 K) x 268 K

       V2  = (0.008 L . K) x 268 K

       V2  = 2.15 L

So the volume at -5.00ºC = 2.15 L

Answer:

figure is attached

Explanation:

When we treat alcohol with H₂SO₄ we get elimination as the major product.

As we can see in the given reaction that in step 1 the lone pair of electrons of oxygen attached to the alcohol make a bond with the hydrogen of H₂SO₄.

In the 2nd step H₂O gets detached from the parent ring which generated a positive charge on the ring.

In the 3rd step elimination of hydrogen from the carbon next to the carbonium carbon results into formation of an alkene.