How would you classify sugar?
a) pure substance-compound.
b) mixture-heterogeneous.
c) pure substance-element.
d) mixture-homogeneous.
e) none of the above.

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:

a

Explanation:

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:

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

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:

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:

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:

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:

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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Strontium is the element in the provided list that has chemical properties similar to magnesium because they both are alkaline earth metals with two valence electrons.

The element in the given list with chemical properties similar to magnesium is strontium. This is because magnesium and strontium both belong to the same group in the periodic table, which is the group of alkaline earth metals.

Elements in the same group share similar chemical properties due to their similar valence electron configurations. Magnesium and strontium, like other alkaline earth metals, have two valence electrons. These two electrons play a crucial role in the chemical reactivity of the elements, including how they bond with other elements.

It's important to note that, despite being in the same group, the reactivity and specific properties vary among the alkaline earth metals. However, the underlying chemical behavior that stems from their valence electron configuration leads to similarities. For example, both magnesium and strontium readily react with water, though strontium's reactivity is somewhat higher.

Final answer:

Strontium is the element with chemical properties similar to magnesium because they both are alkaline earth metals with two valence electrons, belong to the same group in the periodic table, and show similar reactivity patterns.

Explanation:

The element in this list with chemical properties similar to magnesium is strontium (b). This is because the elements that are similar to magnesium would also be in the same group as magnesium in the periodic table. Magnesium is an alkaline earth metal found in Group 2 of the periodic table, which includes beryllium, calcium, strontium, barium, and radium, all known as alkaline earth metals. These elements have two valence electrons and exhibit similar chemical behaviors.

In summary, both magnesium (Mg) and strontium (Sr) are shiny and are good conductors of heat and electricity. Most importantly, the two elements share a common valence electron configuration, which causes them to display similar chemical reactivity patterns, such as forming compounds with a 2+ charge and reacting similarly with other substances.

Answer:

The amount of Cl2 gas left , after the reaction goes to completion is : 139.655 grams

Explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

[tex]Moles=\frac{Mass}{Molar\ mass}[/tex]

[tex]2Al(s)+3Cl_{2}(g)\leftarrow 2AlCl_{3}(g)[/tex]

According  to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

[tex]Moles=\frac{Mass}{Molar\ mass}[/tex]

[tex]Moles=\frac{20.1}{26.98}[/tex]

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

[tex]Moles=\frac{Mass}{Molar\ mass}[/tex]

[tex]Moles=\frac{219}{70.98}[/tex]

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2  mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

Cl2 in grams = 1.964 x 70.90 = 139.655 grams

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:

503.5 kJ

Explanation:

The combustion of reaction of propane, C3H8, is:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)

The enthalpy of the reaction can be calculated by the enthalpy of formation of the substances, which is the enthalpy of the reaction that produces the substances by only its constituents. The values can be found at a thermodynamic table. The standard condition is 25°C and 1 atm, so:

H°f, C3H8(g) = -103.85 kJ/mol

H°f, O2(g) = 0

H°f, CO2(g) = -393.51 kJ/mol

H°f, H2O(l) = -285.83 kJ/mol

So, the enthalpy of the reaction is:

ΔH°rxn = ∑n*H°f products - ∑n*H°f reactants, where n is the coefficient of the substance:

ΔH°rxn = [3*(-393.51) + 4*(-285.83)] - (-103.85)

ΔH°rxn = -2220 kJ/ mol of C3H8

The heat produced is the value of the enthalpy multiplied by the number of moles of the fuel. The molar mass of C3H8 is 44.10 g/mol, so, in 10.0 g:

n = 10.0/44.10

n = 0.2268 mol

So, the heat is:

Q = -2220 * 0.2268

Q = -503.5 kJ

The minus signal indicates that the heat is being lost, so 503.5 kJ of heat is produced.

To calculate the quantity of heat produced when 10.0 g of propane is completely combusted, use the balanced equation for the combustion of propane and the enthalpy of combustion.

When propane (C3H8) is completely combusted in air under standard conditions, it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) vapor. The balanced equation for the combustion of propane is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

To calculate the quantity of heat produced when 10.0 g of propane is completely combusted, we need to use the enthalpy of combustion, which is given as -2,219.2 kJ/mol. To find the heat produced, we first determine the number of moles of propane in 10.0 g, then use the stoichiometry of the balanced equation to convert moles of propane to moles of heat produced, and finally convert the moles of heat to kJ by multiplying by the enthalpy of combustion per mole of propane.

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: 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:

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

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:

a. Heat Energy

b. Temperature

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

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.

Answer:

29.7

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the metal must be gained by the water.  Therefore, heat given off by the metal = −heat taken in by water, or:

The equation used to calculate the quantity of heat energy exchanged in this process is:

Heat stops flowing when the two samples are at the same temperature, so same final temperature of the water will be the final temperature of the metal as well.  

Substitute in the known values for the equation above and rearrange to solve for T.

−(0.622Jg∘C)(73.0 g)(T−105.0∘C)=(4.184Jg∘C)(300. g)(T−27.0∘C)

Simplify by multiplying specific heat and mass.

−(45.406J∘C)(T−105.0∘C)=(1,255.2J∘C)(T−27.0∘C)

Then distribute the heat capacities (calculated in the previous step) to the temperature differences.

−(45.406TfJ∘C)+4,767.63J=(1,255.2TfJ∘C)−33,890J

Combine like terms.

−1,300TfJ∘C=−38,657J

T=29.72∘C

The answer should have three significant figures so round to 29.7∘C.  

=29.72∘C

The final temperature of the water in the mixture is 29.72 ⁰C.

The given parameters;

The final temperature of the water is determined by applying the principle of conservation of energy.

Heat gained by the water = Heat lost by the metal

[tex](mc\Delta t)_{H_2O}\ = (mc\Delta t)_{metal}\\\\300 \times 4.184\times (t - 27) = 73 \times 0.622\times (105 - t) \\\\1255.2 t - 33890.4 = 4,767 - 45.41t\\\\1300.61t = 38657.4\\\\t = \frac{38657.4}{1300.61} \\\\t = 29.72 \ ^0[/tex]

Thus, the final temperature of the water in the mixture is 29.72 ⁰C.

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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:

The answer to your question is below

Explanation:

Reaction

                            C₂H₂ (g) + O₂(g)   ⇒   CO₂ (g)   +  H₂O (g)

                            Reactants         Elements         Reactants

                                    2                       C                       1

                                    2                       H                       2

                                    2                       O                       3

This reaction is unbalance

Reaction balanced

                          2C₂H₂ (g) +   5O₂(g)   ⇒   4CO₂ (g)   +  2H₂O (g)

                            Reactants         Elements         Reactants

                                    4                       C                       4

                                    4                       H                       4

                                   10                       O                      10

Now, the reaction is balanced

a) Calculate the molecular mass of acetylene and water

Acetylene = (12 x 2) + (2) = 26 g

Water = (1 x 2) + (1 x 16) = 18 g

                           2(26) g of Acetylene ---------------  2(18) g of Water

                               113 g  of Acetylene --------------   x

                                x = (113 x (2 x 18)) / 2(26)

                                x = 4068 / 52

                               x = 78. 2 g of water

b)                2 moles of Acetylene ------------  4 moles of carbon dioxide

                   x moles of acetylene ------------  1.10 moles of carbon dioxide

                         x = (1.10 x 2) / 4

                        x = 0.55 moles of acetylene

Answer:

a) 78.19 grams H2O

b) 14.3 grams acetylene

Explanation:

Step 1: Data given

Molar mass of acetylene = 26.04 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

2C2H2 + 5O2 → 4CO2 + 2H2O

Step 3: a. How many grams of water can form if 113g of acetylene is burned?

Calculate moles of acetylene:

Moles = mass / molar mass

Moles = 113.0 grams / 26.04 g/mol

Moles = 4.339 moles

calculate moles of H2O

For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O

For 4.339 moles of acetylene we'll have 4.339 moles H2O

Calculate mass of H2O

Mass H2O = 4.339 moles * 18.02 g/mol

Mass H2O = 78.19 grams H2O

b. How many grams of acetylene react if 1.10 mol of CO2 are produced?

For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O

For 1.10 mol CO2 we need 1.10/2 = 0.55 moles of acetylene

Mass acetylene = 0.55 moles * 26.04 g/mol

Mass acetylene = 14.3 grams acetylene

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:

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:

5118.30485 moles

Explanation:

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

Answer:

c

Explanation:

1 calorie = 4.184J/g×°C

This also happens to be the specific heat capacity of water, which is the amount of energy it takes to raise the temperature of 1mL of water by 1°C

The energy required to change the temperature of 1g of water by 1°C is 1 calorie. This concept is called specific heat.

The energy needed to raise or lower the temperature of 1 g of liquid water by 1°C is referred to as the specific heat of water. The specific heat capacity of water is 1 calorie/gram °C. This means that 1 calorie of heat energy is needed to raise the temperature of 1g of water by 1°C. Therefore, the answer is c. is 1 calorie.

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Answer: A. PV

Explanation: In Boyles Law it is a concept on ideal gases which states the relationship between volume and absolute pressure of the gas is inversely proportional. The relationship can be expressed in PV = k where k is a proportionality constant.

Answer:

A , B and D.

Explanation:

Pressure times volume is a constant.

Answer: The volume of HCl needed is 0.250 L

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

For sodium carbonate:

Molarity of sodium carbonate solution = 0.500 M

Volume of solution = 0.750 L

Putting values in above equation, we get:

[tex]0.500M=\frac{\text{Moles of sodium carbonate}}{0.750}\\\\\text{Moles of sodium carbonate}=(0.500mol/L\times 0.750L)=0.375mol[/tex]

The chemical equation for the reaction of sodium carbonate and HCl follows:

[tex]Na_2CO_3+2HCl\rightarrow 2NaCl+H_2CO_3[/tex]

By Stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 2 moles of HCl

So, 0.375 moles of sodium carbonate will react with = [tex]\frac{2}{1}\times 0.375=0.750mol[/tex] of HCl

Now, calculating the volume of HCl by using equation 1:

Moles of HCl = 0.750 moles

Molarity of HCl = 3.00 M

Putting values in equation 1, we get:

[tex]3.00M=\frac{0.750mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.750mol}{3.00mol/L}=0.250L[/tex]

Hence, the volume of HCl needed is 0.250 L

Answer:

The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol

Explanation:

Fractional distillation is a separation process based on difference in boiling point of two compounds.

1-chlorobutane is a polar aprotic molecule due to presence of polar C-Cl bond. Hence  dipole-dipole intermolecular force exists in 1-chlorobutane as a major force.

1-butanol is a polar protic molecule. Hence dipole-dipole force along with hydrogen bonding exist in 1-butanol.

Therefore intermolecular force is stronger in 1-butanol as compared to 1-chlorobutane.

So, boiling point of 1-butanol is much higher than 1-chlorobutane.

Hence mixture of 1-chlorobutane and 1-butanol can be separated by fractional distillation based on difference in boiling point.

So, option (D) is correct.

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: