Se mezclan 20 gramos de agua (1cal/g°C) a 40 °C con 15 gramos de alcohol (0,58cal/g°C) a 30 °C. ¿Cuál ha sido la temperatura de equilibrio térmico?

Answer:

"Symbolic representation of a chemical reaction in the form of symbols and formulae"

Answer:

A chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulas.

Explanation:

Example- Zinc metal reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas. This equation is written as: Zn+H2SO4→ZnSO4+H2.

Answer:

As we know that

At STP

1 mole of any gas occupies 22.4L  

so

15.7L / 22.4L/mole = 0.7moles Cl2  

0.7moles x 6.022x10^23molecules/mole = 4.22x10^23molecules

hope it helps

The final temperature of the aluminum engine part is approximately 284.19 degrees Celsius.

To determine the final temperature of the aluminum engine part, we can use the equation:

q = m × c × ΔT

Where:

q is the heat absorbed (in J)

m is the mass of the aluminum part (in g)

c is the specific heat capacity of aluminum (in J/g·K)

ΔT is the change in temperature (in K)

First, let's convert the mass of the aluminum part from grams to kilograms:

295 g = 0.295 kg

We are given the heat absorbed as 75.0 kJ, which is equal to 75,000 J.

Now, let's rearrange the equation to solve for ΔT:

ΔT = q / (m × c)

Substituting the values:

ΔT = 75,000 J / (0.295 kg × 0.900 J/g·K)

Calculating ΔT:

ΔT = 271.19 K

To find the final temperature, we need to add the change in temperature to the initial temperature:

Final temperature = Initial temperature + ΔT

Final temperature = 13.00°C + 271.19 K

Converting back to Celsius:

Final temperature = 284.19°C

Therefore, the aluminum engine part's ultimate temperature is approximately 284.19 degrees Celsius.

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Answer: 35.078 to 3 significant figures

Explanation:

Solving;

34.56 + 42.235 * 0.0123 - 0.001

We solve the multiplication first according to the BODMAS rule.

So,

42.235 * 0.0123 = 0.51949

Doing the addition,

34.56 + 0.51949 - 0.001

This gives;

35.079 - 0.001

And finally the subtraction gives;

= 35.078

Answer:

the state of matter changed

Explanation:

Answer:

A) A precipitate is formed

Explanation:

occurs in aqueous solution when two ions bond together to form an insoluble salt, which is known as the precipitate.

The statement that supports the claim that mass is conserved in a chemical reaction is the law of conservation of mass, which states that the total mass of the reactants is equal to the total mass of the products.

According to the law of conservation of mass, which is a fundamental principle in chemistry, mass is conserved in a chemical reaction. This means that the total mass of the substances involved in the reaction remains constant before and after the reaction takes place. In other words, no mass is gained or lost during a chemical reaction. The atoms in the reactants rearrange themselves to form new substances, but the total number of atoms remains the same. Therefore, the mass of the reactants is equal to the mass of the products. This principle has been experimentally verified and is a cornerstone of chemical reactions.


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☁️ Answer ☁️

Yes.

"In so doing, he became the first scientist to explain the behavior of atoms in terms of the measurement of weight. He also uncovered the fact that atoms couldn't be created or destroyed. Dalton's theory additionally examined the compositions of compounds, explaining that the tiny particles (atoms) in a compound were compound atoms"

Hope it helps.

Have a nice day noona/hyung.

When light is incident on the ion, wavelengths around 500 nm will be absorbed, and the remaining wavelengths will be transmitted or reflected. This provides insight into the electronic structure of the titanium(III) ion and its interaction with light.

The titanium(III) ion, [Ti(H2O)6]3+ in water, absorbs light that is centred around 500 nm. The colour of light absorbed by the ion can be inferred from the visible spectrum of electromagnetic radiation, which ranges from about 400 to 700 nm. Wavelengths around 400-450 nm appear violet, 450-495 nm blue, 495-570 nm green, 570-590 nm yellow, 590-620 nm orange, and 620-700 nm red. Since the titanium(III) ion absorbs light around 500 nm, we can infer that it absorbs light in the green portion of the visible spectrum. Therefore, the titanium(III) ion appears to be a green color.

The absorption of light by compounds can provide insight into the colour of the compound. When light is incident on a compound, certain wavelengths are absorbed, and the remaining wavelengths are transmitted or reflected. The absorbed wavelengths of light depend on the electronic structure of the compound. In the case of the titanium(III) ion, [Ti(H2O)6]3+, it is observed that light centered around 500 nm is absorbed. This means that the electronic structure of the ion allows it to absorb light in the green portion of the visible spectrum. Since the visible spectrum ranges from approximately 400 to 700 nm, it can be inferred that the titanium(III) ion appears to be green in colour. Therefore, when light is incident on the ion, wavelengths around 500 nm will be absorbed, and the remaining wavelengths will be transmitted or reflected. This provides insight into the electronic structure of the titanium(III) ion and its interaction with light.

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The reaction between 1,3-cyclohexadiene and maleic acid (cis-2-butene-dioic acid) leads to the formation of a Diels-Alder adduct.

The product formed is a bicyclic compound resulting from the cycloaddition reaction between 1,3-cyclohexadiene and maleic acid. The reaction involves the formation of a new six-membered ring fused with the cyclohexene ring of 1,3-cyclohexadiene.

The reaction proceeds via the concerted [4 + 2] cycloaddition mechanism, where the diene (1,3-cyclohexadiene) reacts with the dienophile (maleic acid) to form a new sigma bond and two new pi bonds. The resulting adduct exhibits a fused ring system with a double bond in the newly formed six-membered ring.

The specific arrangement of substituents and stereochemistry in the product can vary depending on the orientation of the reactants and the conditions of the reaction. It is important to consider the regioselectivity and stereochemistry when drawing the complete structure of the product.

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The density of the metal is 2.7 g/cm³, which can be used to identify the metal since different elements have different densities.

The given mass of the metal is 27g and its volume is 10 cm³.

The identity of the metal can be found out by using the formula of density, which is Density = Mass/Volume.

Density is a physical property of matter that is the amount of mass per unit volume. By rearranging the formula of density, we can determine the identity of the metal by finding its density which is given by mass divided by volume.

Identity of a metal that has a mass of 27g and a volume of 10 cm³ can be determined by using the formula of density which is Density = Mass/Volume.  

Here is the solution:Given,Mass of the metal = 27gVolume of the metal = 10 cm³

Density of the metal is given by the formula:Density = Mass/Volume

Substituting the given values in the formula:Density = 27g/10 cm³ = 2.7 g/cm³

The periodic table can be used to match the density with known elements and thus determine the identity of the metal.

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The moles of propane consumed in the reaction is approximately 4.00 moles. Option B is correct.

The balanced equation for the reaction is;

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

According to the stoichiometry of the reaction, for every 1 mole of C₃H₈, 5 moles of O₂ are required for complete combustion.

Given that you start with 4.00 moles of C₃H₈ and 4.00 moles of O₂, we need to determine the limiting reactant to determine the moles of propane consumed.

For C₃H₈:

moles of C₃H₈ = 4.00 moles (given)

For O₂;

moles of O₂ = 4.00 moles (given)

To determine the limiting reactant, we need to compare the mole ratios of the reactants to the stoichiometry of the balanced equation.

For C₃H₈;

moles of C₃H₈ used = 4.00 moles

For O₂;

moles of O₂ required = 5 moles of O₂ per 1 mole of C₃H₈

moles of O₂ required = 5 × 4.00 moles = 20.00 moles

Comparing the moles of C₃H₈ used (4.00 moles) and the moles of O₂ required (20.00 moles), we can see that the moles of C₃H₈ used are less than the moles of O₂ required. This means that C₃H₈ is the limiting reactant.

Therefore, the moles of propane is 4.00 moles.

Hence, B. is the correct option.

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--The given question is incomplete, the complete question is

"Consider the reaction below. If you start with 4.00 moles of C₃H₈ (propane) and 4.00 moles of O₂, how many moles of propane would be consumed? C₃Hg(g) + 5 O₂(g) + 3 CO₂(g) + 4H₂O(g) A) 3.20 B) 4.00 C) 20.0 D) 2.00 E) 0.800."--

The correct statement which are true about a galvanic cell that has an E value of 1.00V are The reaction is spontaneous and It has a negative ΔG°. Hence, the correct option is e. The species produced by a beta emission by magnesium-27 is no specific isotope or element. Hence, the correct option is e.

A galvanic cell with an E value of 1.00V indicates that the cell reaction is spontaneous. The E value represents the cell potential, and a positive E value indicates that the reaction will proceed spontaneously in the forward direction.

At equilibrium, the cell potential would be zero (E = 0), but the given cell has an E value of 1.00V. Therefore, statement II is incorrect. The equilibrium constant (K) is not necessarily equal to 1 in a galvanic cell.

The ΔG° (standard Gibbs free energy change) is related to the E value of the cell by the equation: ΔG° = -nFΔE°, where n is the number of moles of electrons transferred and F is the Faraday constant. Since the E value is positive (1.00V), the ΔG° will be negative, indicating that the cell reaction is thermodynamically favorable. Therefore, statement III is true.

Therefore, the correct answer is e. I and III.

The beta emission by magnesium-27 (27Mg) would result in the production of an electron (β- particle). Therefore, the correct answer is e. None. No specific isotope or element is produced by the beta emission of magnesium-27.

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Carbon dioxide is considered to be a greenhouse gas because it absorbs and emits infrared radiation.

Greenhouse gases like carbon dioxide trap heat in the atmosphere, leading to the warming of the Earth's surface and atmosphere.Carbon dioxide is a greenhouse gas, meaning that it can absorb and emit infrared radiation. Carbon dioxide and other greenhouse gases in the atmosphere act as a type of insulation, trapping heat and keeping the Earth's surface and atmosphere warm.A greenhouse gas is a gas that can absorb and emit infrared radiation. The greenhouse effect occurs when certain gases, such as carbon dioxide and water vapor, trap heat in the Earth's atmosphere. As more greenhouse gases are released into the atmosphere, the Earth's temperature increases, leading to climate change.

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Answer: Ik its too late but the answer is yeast.

Explanation: Sugar is yeasts food. If you mix sugar with yeast, and put it in something such as bread, a bottle with a balloon over it, or etc. It will cause it to rise.

Answer:

150,000 Joules

Explanation:

KE= 1/2 *mv^2

1/2 * 120 *50^2

60*2500

150,000 Joules

Hope this helps :)

Answer:

The correct answer is Choice C.

(Phenol red indicator.)

Explanation:

Phenol red indicator is an organic dye that acts as an acid-base reaction indicator in the pH range of 6.8 to 8.4.

In acidic solutions, the color is purple, and in simple solutions, the color is red.

Hope this helps!

Please mark me as Brainlinieast.

Answer:

phenol red indicator

Explanation:

The free energy change (ΔG°) for the reaction of 2.46 moles of H2S(g) at standard conditions and 298K is approximately 252 kJ.G° for this reaction is expected to be less than zero.

The equilibrium constant (Keq) for a reaction is related to the standard Gibbs free energy change (ΔG°) through the equation:

ΔG° = -RTln(Keq)

Where:

ΔG° is the standard Gibbs free energy change

R is the gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin (298 K)

ln is the natural logarithm

Given that Keq = 8.47×10^(-37), we can calculate the standard Gibbs free energy change (ΔG°) using the equation above:

ΔG° = - (8.314 J/(mol·K)) * (298 K) * ln(8.47×10^(-37))

Calculating this value gives us:

ΔG° = - (8.314 J/(mol·K)) * (298 K) * (-87.24)

ΔG° ≈ 2.52 × 10^5 J/mol

Converting the units to kilojoules (kJ/mol), we get:

ΔG° ≈ 252 kJ/mol

The free energy change (ΔG°) for the reaction of 2.46 moles of H2S(g) at standard conditions and 298K is approximately 252 kJ. Since the value of ΔG° is positive, it indicates that the reaction is not spontaneous and requires an input of energy to proceed.

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Answer:

Thank youuuuu bestie, you really pulling through

Answer:

yessss

Explanation:

Answer:

6.6M

Explanation:

C=concentration

V=Volume

C1V1=C2V2

rearrange formula to find C2

(C1V1)/V2= C2

(12M x 550mL)/1000mL= C2

6600M*mL*/1000*mL* =C2 (cancel units)

6.6M=C2

Answer:

Yes, the combustion of hydrogen is spontaneous from the given conditions.

Explanation:

Let look at the reaction:

[tex]2H_2_{(g)} + O_2_{(g)} \to 2H_2O_{(g)}[/tex]

[tex]K_p = \dfrac{[P(H_2O)]^2}{[P(H_2)^2\times P(O_2)]} ----- (1)[/tex]

The total pressure of the tank = 1 atm

[tex]\mathtt{Pressure \ of \ the \ gas = moles \ fraction \of \ gas \times total \ pressure}[/tex]

Thus, pressure of [tex]O_2[/tex] gas = [tex]\dfrac{9.38 \times 10^{-3} }{(9.38 \times 10^{-3} + 2.46 \times 10^{-8} + 1.79 \times 10^{-4}} \times 1 atm[/tex]

[tex]P(O_2) = 0.9813 atm[/tex]

Pressure of [tex]H_2O[/tex] gas =[tex]\dfrac{1.78 \times 10^{-4}}{9.38 \times 10^{-3} + 2.46 \times 1)^{-8} + 1.78 \times 10^{-4}} \times 1 \ atm[/tex]

Pressure of H_2 gas = [tex]2.573 \times 10^{-6} \ atm[/tex]

From equation (1);

[tex]K_p = \dfrac{(0.0187)^2}{(2.573 \times 10^{-6})^2 \times (0.913) } \\ \\ K_p = 5.383 \times 10^7[/tex]

So, applying the equation:

[tex]\Delta G^0 = -RT In K_p \\ \\ = -8.314 \times (300)K \times IN(5.383 \times 10^7) \\ \\ = -45880.109 J \\ \\ = -45.800 \ kJ\\ \\[/tex]

Thus;

[tex]\Delta G^0 < 0[/tex]

Thus, Yes the combustion of hydrogen is spontaneous from the above conditions. From thermodynamics, we realize that assuming ∆G is negative, the reaction is spontaneous. Again we understand that ∆G=∆H-T∆S. However, for combustion, water will definitely be formed.

Finally, we can conclude that;

∆H is negative;

∆S is positive; &

∆G is also negative.

Thus, from above, we conclude combustion of hydrogen is spontaneous.

The combustion of hydrogen gas with the oxygen gas for the formation of the water vapor is a spontaneous reaction.

To know about the condition that the given reaction is spontaneous or not we will use the below equation and must get the negative value:

ΔG° = -RTlnKp, where

R = universal gas constant

T = temperature

Kp = partial pressure constant

Firstly we have to calculate the value of Kp.

Given chemical reaction is:

2H₂(g) + O₂(g) → 2H₂O(g)

Value of Kp for the given reaction is:

Kp = [p(H₂O)]² / [p(H₂)]².[p(O₂)]

Partial pressure will be calculated as:

p = mole fraction × total pressure

Given that total pressure of gas = 1atm

p(H₂O) = (1.79×10⁻⁴ / 1.79×10⁻⁴+2.46×10⁻⁸+9.38×10⁻³) × 1 = 0.0187

p(H₂) = (2.46×10⁻⁸ / 1.79×10⁻⁴+2.46×10⁻⁸+9.38×10⁻³) × 1 = 2.573 × 10⁻⁶

p(O₂) = (9.38×10⁻³ / 1.79×10⁻⁴+2.46×10⁻⁸+9.38×10⁻³) × 1 = 0.913

Putting all these values on the equation of Kp, we get

Kp = (0.0187)² / (2.573 × 10⁻⁶)²(0.913) = 5.383 × 10⁷

Now putting values in the first equation we get,

ΔG° = -(8.314)(300)ln(5.383 × 10⁷)

ΔG° = -45,800 kJ

Hence, given reaction is spontaneous.

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Among the salts listed, ZnCO3 (zinc carbonate) and Ba3(PO4)2 (barium phosphate) are the salts that will be substantially more soluble in acidic solution than in pure water.

In the case of ZnCO3, it is an insoluble salt in pure water due to the carbonate ion's basic nature. Carbonate ions (CO3^2-) react with water molecules to form bicarbonate ions (HCO3-) and hydroxide ions (OH-) in an equilibrium reaction. The presence of an acidic solution would shift this equilibrium toward the reactant side, favoring the formation of CO3^2- ions and increasing the solubility of ZnCO3. Similarly, Ba3(PO4)2, which is barium phosphate, is insoluble in pure water. Phosphate ions (PO4^3-) have a basic nature and tend to form insoluble salts with many cations. In an acidic solution, the excess of hydrogen ions (H+) would react with phosphate ions, forming dihydrogen phosphate ions (H2PO4-) or monohydrogen phosphate ions (HPO4^2-). This reaction reduces the concentration of phosphate ions, decreasing the formation of insoluble Ba3(PO4)2 and enhancing its solubility. On the other hand, ZnS (zinc sulfide), BiI3 (bismuth triiodide), and AgCN (silver cyanide) do not show a significant change in solubility in an acidic solution compared to pure water. Their solubilities are primarily governed by factors such as the lattice energy and the ion-ion interactions within the crystal lattice, which are less influenced by changes in pH.

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Answer:

Scientists study Earth's interior by studying

seismic waves. Seismic waves are waves of energy that travel

through Earth.

Explanation:

The correct expression relating the rates of consumption of reactants and the rates of formation of products for the given reaction is A[[tex]LaCl_{3}[/tex]] / 2Δt = A[La_{2}(CO_3){3}] / 3Δt = A[[tex]Na_{2}CO_{3}[/tex]] / Δt.

The balanced chemical equation for the reaction is 2LaCl_{3}+ 3[tex]Na_{2}CO_{3}[/tex]→ La_{2}(CO_3){3} + 6NaCl. To determine the expressions relating the rates of consumption of reactants and the rates of formation of products, we can use the stoichiometric coefficients from the balanced equation. According to the stoichiometry of the reaction, for every 2 moles of [tex]LaCl_{3}[/tex]consumed, 1 mole of La_{2}(CO_3){3}is formed. Therefore, the expression relating the rates of consumption of LaCl3 and formation of La_{2}(CO_3){3}is A[LaCl_{3}] / 2Δt = A[[tex]La_{2}(CO_3){3}[/tex]] / Δt. Similarly, for every 3 moles of [tex]Na_{2}CO_{3}[/tex]consumed, 1 mole of La_{2}(CO_3){3} is formed. Therefore, the expression relating the rates of consumption of [tex]Na_{2}CO_{3}[/tex] and formation ofLa_{2}(CO_3){3}is A[] / 3Δt = A[La_{2}(CO_3){3}] / Δt.

From the balanced equation, we can also see that 6 moles of NaCl are formed for every 2 moles of LaCl3 consumed. However, the question does not provide options involving the rate of consumption of NaCl. In conclusion, the correct expression relating the rates of consumption of reactants and the rates of formation of products for the given reaction is A[LaCl_{3}] / 2Δt = A[La_{2}(CO_3){3}] / 3Δt = A[[tex]Na_{2}CO_{3}[/tex]] / Δt.

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Balanced half-reactions for the given redox reaction are Reduction: Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l), Oxidation: Zn(s) → Zn2+(aq) + 2 e-. As the reaction proceeds, Zn is oxidized and Cr2O72- is reduced. It is a redox reaction because it involves both oxidation and reduction.

In the half-reaction of oxidation, zinc atoms donate two electrons to become zinc ions. Therefore, oxidation occurs at the anode, which produces electrons that flow into the external circuit. In the reduction half-reaction, Cr2O72- accepts six electrons and seven hydrogen ions to become Cr3+ and water. Therefore, reduction occurs at the cathode, which consumes the electrons produced by the anode. Therefore, the overall balanced redox reaction is Cr2O72-(aq) + 14 H+(aq) + 3 Zn(s) → 2 Cr3+(aq) + 7 H2O(l) + 3 Zn2+(aq). The reaction can be explained as follows: zinc atoms donate two electrons to each hydrogen ion present in the acidic solution to form hydrogen gas, which causes slow corrosion of the zinc electrode. The dichromate ion acts as the oxidizing agent by accepting electrons from the zinc electrode to form zinc ions.

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The concentrations of the species in the 1.40 M NaCH₃COO solution are (Na⁺) = 1.40 M, (OH⁻) = 0 M, (H₂O⁺) = 0 M, (CH₃COOH⁻) = 0.00502 M, and (CH₃COOH) = 0.00502 M.

To calculate the concentrations of the species in a 1.40 M NaCH₃COO (sodium acetate) solution, we need to consider the dissociation of sodium acetate and the ionization of acetic acid.

The dissociation of sodium acetate (NaCH₃COO) can be represented as follows

NaCH₃COO → Na⁺ + CH₃COO⁻

Since sodium acetate fully dissociates in water, the concentration of Na⁺ will be the same as the initial concentration of sodium acetate: 1.40 M.

The ionization of acetic acid (CH₃COOH) can be represented as follows

CH₃COOH ⇌ H⁺ + CH₃COO⁻

The ionization constant for acetic acid (Ka) is given as 1.8×10⁻⁵

Let's denote the initial concentration of acetic acid as x M. At equilibrium, the concentration of H⁺ and CH₃COO⁻ will also be x M.

Using the ionization constant expression for acetic acid, we can write

Ka = [H⁺][CH₃COO⁻] / [CH₃COOH]

1.8×10⁻⁵ = x * x / (1.40 - x)

Since the value of x is expected to be small compared to 1.40, we can approximate the denominator as 1.40.

1.8×10⁻⁵ = x² / 1.40

Rearranging the equation

x² = 1.8×10⁻⁵ × 1.40

x² = 2.52 × 10⁻⁵

Taking the square root of both sides

x = √(2.52×10⁻⁵)

x ≈ 0.00502 M (rounded to four decimal places)

Now we can calculate the concentrations of the species

(Na⁺) = 1.40 M (same as the initial concentration of sodium acetate)

(OH⁻) = 0 M (since sodium acetate and acetic acid do not provide OH⁻ ions)

(H₂O⁺) = 0 M (water does not provide H⁺ ions)

(CH₃COOH⁻) = 0.00502 M

(CH₃COOH) = 0.00502 M

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